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Lebesgue measure on the real numbers extends interval length: $\lambda ([a,b])=b-a$ for $a\le b$. Its restriction to the Borel sigma-algebra is the unique translation-invariant sigma-finite measure on that sigma-algebra normalized by $\lambda ((0,1])=1$. The full Lebesgue measure is the completion of a measure of this restriction; its domain is the Lebesgue measurable sets. Translation invariance follows by comparing a translated Borel measure on half-open intervals and applying the uniqueness theorem for measures. For the construction and interval properties, see MIT lecture 8.

Every singleton has Lebesgue measure zero: $\{x\}\subseteq [x-\epsilon ,x+\epsilon ]$ gives $0\le \lambda (\{x\})\le 2\epsilon$ for every $\epsilon >0$. Enumerating the countable set $\mathbb{Q}\cap [0,1]$ and applying countable additivity yieldsThus a dense set can be a null set, and a union of null sets indexed by an uncountable set can have positive Lebesgue measure. The distinction explains both the Dirichlet function example and why almost everywhere is weaker than everywhere.

On a measure space of positive total measure, the L-infinity norm of an essentially bounded measurable function is $\Vert f{\Vert }_{\infty }={\mathrm{esssup}}_{\mu }|f|$. The absolute value matters: a constant function $f=-3$ has L-infinity norm three, even though its essential supremum is $-3$. The norm belongs to the equivalence class in Lp space; different representatives equal almost everywhere give the same answer. ^[ref].

For the function $g$ in essential infimum and essential supremum, $\Vert g{\Vert }_{\infty }=1$ under Lebesgue measure, whereas its supremum norm is four. Replacing the exceptional values by the ordinary values $x$ leaves every finite Lp norm unchanged as well: $\Vert g{\Vert }_p=({\int }_0^1{x}^{p}dx{)}^{1/p}=(p+1{)}^{-1/p}$. This tends to one as $p\to \infty$. That agreement explains why the endpoint with exponent $p=\infty$ uses an essential bound rather than a pointwise bound.

A finite probability space makes the distinction between zero and small measure explicit. Let $f(a)=1$, $f(b)=3$, and $f(c)=10$. Assign weights $\mu (a)=1/2-\epsilon$, $\mu (b)=1/2$, $\mu (c)=\epsilon$, where $0\le \epsilon <1/2$. For $1\le p<\infty$,When $\epsilon =0$, the singleton $\{c\}$ is a null set and $\Vert f{\Vert }_{\infty }=3$. For every $\epsilon >0$, however small, $\Vert f{\Vert }_{\infty }=10$. For example, $\epsilon =0.001$ gives $\Vert f{\Vert }_1=2.009$ and $\Vert f{\Vert }_2=\sqrt{5.099}$, yet the large-$p$ Lp norms approach ten. The ordinary pointwise supremum norm is ten in both cases.

Why can a rare value dominate? For a bounded function on a probability space, an almost everywhere bound $|f|\le M$ gives $\Vert f{\Vert }_p\le M$. If $0<t<\Vert f{\Vert }_{\infty }$, the set $A_t=\{|f|>t\}$ has positive probability, so $\Vert f{\Vert }_p\ge t\mu (A_t{)}^{1/p}$. As $p\to \infty$, the factor $\mu (A_t{)}^{1/p}$ tends to one. These two bounds squeeze the limit to $\Vert f{\Vert }_{\infty }$. The probability measure assumption keeps total measure finite; this explanation does not assert the same finite-$p$ integrability on arbitrary infinite-measure spaces.

An Lp norm measures the size of a measurable function by averaging a power of its absolute value with respect to a specified measure. For $1\le p<\infty$,when the integral is finite. The exponent $p$ controls how strongly large values influence the result. The Lp space identifies functions equal almost everywhere, so this becomes a norm on the corresponding equivalence classes. For $p=\infty$, use the essential supremum of $|f|$, as defined by the L-infinity norm. ^[ref].

For a two-coordinate vector $(3,4)$, treat the coordinates as a function on two inputs with counting measure. Then $\Vert (3,4){\Vert }_1=3+4=7$, $\Vert (3,4){\Vert }_2=\sqrt{9+16}=5$, and $\Vert (3,4){\Vert }_{\infty }=4$. The middle case is the Euclidean norm; the first adds the coordinate magnitudes, and the last takes their largest value. These are different answers to different questions about size, all satisfying the triangle inequality.

If instead each input has probability $1/2$, the same values give $\Vert f{\Vert }_1=3.5$ and $\Vert f{\Vert }_2=\sqrt{12.5}$, while $\Vert f{\Vert }_{\infty }=4$ remains unchanged. Thus the measure is not an optional label. On a probability space, the finite Lp norms are nondecreasing in $p$ whenever they are finite, by Hölder's inequality; the unnormalized two-coordinate example above decreases instead. Statements comparing different exponents must specify their measure convention.

Why are these sizes useful? If $e$ is an error function, its L1 norm measures aggregate absolute error, its L2 norm measures the square root of aggregate quadratic loss, and its L-infinity norm bounds error almost everywhere. Concentrating a large error on a set of small positive measure can give a modest finite Lp norm but a large L-infinity norm. The three-point example and diagram in L-infinity norm show this explicitly, including why a null set behaves differently from a rare event.

Measure theory studies measures, which assign sizes to measurable sets, and the Lebesgue integral, which accumulates values weighted by those sizes. The same framework describes Lebesgue measure on the real numbers, counting measure on discrete collections, and probability measures on possible outcomes. The choice of measure determines what counts as large; the choice of sigma-algebra determines what can be measured.

A striking example separates size from proximity: the rational numbers in $[0,1]$ form a dense set with Lebesgue measure zero, although the whole interval has Lebesgue measure one. This distinction makes almost everywhere statements useful: exceptional null sets can be ignored in a Lebesgue integral even when they meet every open subinterval. The monotone convergence theorem and dominated convergence theorem then give precise conditions for passing a limit through an integral.

A historical starting point is Lebesgue's 1902 thesis, ^[ref]. Its title connects integration with the geometric task of assigning sizes.

If zero is a regular value of the joint smooth aggregate excess demand $Z(p,t)$, almost every parameter gives a regular economy. The regular level set theorem makes the zero set a manifold. Its projection to parameter space fails to have full rank exactly where the derivative with respect to prices is singular; Sard theorem makes these exceptional parameter values a null set.

The points belonging to infinitely many $E_n$ form their limit superior of sets:For every $N$, countable subadditivity givesThus $\boxed{m(G)=0}$. Outside this null set, each point belongs to only finitely many $E_n$, so its indicator functions are eventually zero. Therefore $\boxed{{\chi }_{E_n}(x)\to 0\text{ almost everywhere}}$. This is the summable-measure argument behind the first Borel-Cantelli lemma; no independence assumption is needed.

Assume the uniform estimate obtained in the geometric setting of part (i). The limit $g$ is bounded and continuous on the finite-alphabet symbolic space, so $p(x)=e^{g(E(x))}$ is measurable, positive, and bounded. Possible duplicate itineraries at partition endpoints do not affect its Lebesgue measure density.

For a level-$n$ interval ${\Delta }_w$, the estimate givesFix a level-$m$ cylinder $C_v$ and subdivide its interval into level-$n$ intervals, with $n\ge m$. Integrating these inequalities and summing the subcylinder masses yieldsLetting $n\to \infty$ proves the exact equality ${\int }_{{\Delta }_v}pdx=\nu (C_v)$. The same argument on the entire interval gives $\int p=1$. Shrinking cylinders generate the interval Borel sigma-algebra, so the Monotone class theorem identifies $p(x)dx$ with the pushforward ${\pi }_{\ast }\nu$.

The symbolic Gibbs measure in this setting is shift invariant, and coding satisfies $T\circ \pi =\pi \circ \sigma$ away from the endpoint exceptions. These exceptions form a null set: bounded $p$ and shrinking cylinders give zero mass to individual endpoints. Therefore, for every Borel set $B$,Consequentlyis an absolutely continuous invariant measure. This implication uses the uniform cylinder estimate directly and does not rely on naming an invariance theorem in place of the requested argument.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.

Normalize the riskless financial asset to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain random vector is $Y=S_1/b-S_0$. A zero-cost portfolio holding $h\in {\mathbb{R}}^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $bh\cdot Y$. Thus an arbitrage meansAn equivalent martingale measure is a probability measure $Q$ with the same null sets as $P$, under which $Y$ is integrable and ${\mathbb{E}}_{Q}Y=0$. The discounted risky prices are then martingales over the single trading period.

If such $Q$ exists, a nonnegative terminal gain has expectation zero under $Q$. A nonnegative random variable with zero expectation vanishes almost surely: on $\{X\ge 1/n\}$ its expectation is at least $\mathbb{P}(X\ge 1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding arbitrage.

For the converse we prove the needed positive-density separation proof of the one-period asset-pricing theorem. First make the gain integrable without changing its null sets. Define an equivalent probability measure $P_0$ byLet $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^{\perp }$ be its orthogonal complement. Taking a finite basis of $N$ shows $Y\in L$ almost surely. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the positive-weight expectation conePositive combinations of weights show that $C$ is a convex cone. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it toThese weights remain bounded and strictly positive, whileFor every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive probability; otherwise $v\in N\cap L$. ConsequentlyThe matrix is finite because $\Vert Y{\Vert }^2/(1+\Vert Y\Vert )\le \Vert Y\Vert$. The kernel of $A_Z$ on $L$ is zero, so it defines an invertible matrix on the finite-dimensional vector space $L$. For any sufficiently small $\delta >0$, every $y$ with $\Vert y\Vert <\delta /\Vert A_Z^{-1}\Vert$ is the image of a vector in the radius-$\delta$ ball, since $\Vert A_Z^{-1}y\Vert <\delta$. Hence the perturbations give a neighborhood of zero, proving relative openness of $C$.

If $0\notin C$, the hyperplane separation theorem gives a nonzero $h\in L$ with $h\cdot c\ge 0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=1_A+\epsilon$. Taking the limit as $\epsilon \downarrow 0$ yields ${\mathbb{E}}_{P_0}[(h\cdot Y)1_A]\ge 0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an arbitrage, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies ${\mathbb{E}}_{P_0}[ZY]=0$. Normalize it by $dQ/d{P}_0=Z/{\mathbb{E}}_{P_0}Z$. This gives an equivalent martingale measure with integrable gains, establishing the fundamental theorem of asset pricing here without any further existence theorem.

Now retain only the $m$ positive-probability states of the finite probability space. Let $V\subseteq {\mathbb{R}}^m$ be the linear span of the discounted terminal asset payoffs; it contains the constant vector $1$. A complete market means $V={\mathbb{R}}^m$. An equivalent martingale measure is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V={\mathbb{R}}^m$, the indicator functions of individual states are attainable, so these prices determine every $q_j$ uniquely.

If $V$ is proper, choose $0\ne v\in V^{\perp }$. Since $1\in V$, ${\sum }_j{v}_j=0$. For sufficiently small $\epsilon >0$, both $q+\epsilon v$ and $q-\epsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct equivalent martingale measures. Therefore market completeness is equivalent to uniqueness of the equivalent martingale measure.

Here is the one-parameter Kolmogorov continuity theorem. Suppose a real-valued stochastic process on $[0,1]$ satisfies, for some $p>0$, $q>0$ and finite $C$,Then it has a continuous modification of a stochastic process whose paths are Hölder continuous of every positive exponent $\gamma <q/p$. A modification means equality almost surely at each fixed time, not an a priori simultaneous equality at all times. Other compact time intervals follow by rescaling; the vector-valued version uses a norm in place of absolute value.

For the proof fix $0<\gamma <q/p$ and use the dyadic grids $D_n=\{j{2}^{-n}:0\le j\le 2^n\}$. Let $D_n^{\ast }$ be the largest adjacent increment on this grid. The union bound and Markov inequality giveThe right-hand side is summable. The Borel-Cantelli first lemma shows that, outside a null set, an almost-surely finite random constant $K$ can absorb the finitely many early grids so that

We now use dyadic increment chaining. For two distinct dyadic points $s,t$, choose $n$ with $2^{-(n+1)}<|t-s|\le 2^{-n}$. Approximate them from below by $s_j,t_j\in D_j$, for $j\ge n$. Each refinement moves by zero or one grid step, while $s_n,t_n$ are at most two grid steps apart. The sequences eventually equal the original dyadic points. Telescoping giveswhere a deterministic finite $C_{\gamma }$ absorbs the geometric series and $2^{-n}<2|t-s|$. Thus the sample function on the dense union of dyadic grids has a unique Hölder-continuous extension to $[0,1]$. Define this extension to be $\tilde{X}$ on the probability-one event, and set it to zero on the exceptional null set.

For a fixed $t$, take dyadic $t_j\to t$. The moment condition gives $X_{t_j}\to X_t$ in probability, while the constructed extension gives $X_{t_j}\to {\tilde{X}}_t$ almost surely. Uniqueness of a probability limit yields ${\tilde{X}}_t=X_t$ almost surely. This proves the required modification property. Performing the argument for a countable increasing sequence of exponents tending to $q/p$ gives one version with all smaller positive Hölder exponents, since its extensions agree on the dyadic points.

For Brownian motion, Gaussian increments give, for any integer $r\ge 2$,Use $p=2r$ and $q=r-1$. The theorem gives every exponent $\alpha <(r-1)/(2r)=1/2-1/(2r)$. For any $0<\alpha <1/2$, choose $r$ large enough. Intersect probability-one events over integer $r$ and compact intervals $[0,N]$ to obtainThe usual continuous Brownian version has the same property: its continuous paths agree with the constructed version on a countable dense set and hence everywhere, outside one null set. The Hölder constant depends on the compact interval. This is the precise local meaning of Brownian Hölder regularity; it is not a uniform Hölder bound on the entire unbounded time axis.

Normalize the second price to one and write $p>0$ for the first price and $t>0$ for the varying endowment. Let $Z(p,t)$ be first-good aggregate excess demand. Under Walras law, the second excess-demand coordinate equals $-pZ(p,t)$, so one market-clearing equation suffices. A regular exchange economy at fixed $t$ has $Z_p(p,t)\ne 0$ at every equilibrium.

A sufficient joint condition is that zero be a regular value of the smooth map $Z:(0,\infty {)}^2\to \mathbb{R}$:To prove the almost-everywhere conclusion, the regular level set theorem makes $M=Z^{-1}(0)$ a smooth one-dimensional manifold. Project $M$ to the parameter axis by $\pi (p,t)=t$. Its tangent space consists of vectors $(h,k)$ with $Z_{p}h+Z_{t}k=0$. If $Z_p\ne 0$, every $k$ is possible, so the derivative of $\pi$ is surjective. If $Z_p=0$, the joint condition gives $Z_t\ne 0$, forcing $k=0$; the point is critical for $\pi$.

Thus the parameters with a singular equilibrium are exactly the critical values of this projection. Sard theorem makes that set a null set. Therefore almost every parameter gives a regular economy. No claim of existence at every parameter is needed for this sufficient regularity criterion.

We first prove invariance, before using any assertion that requires it. Write $A_n\phi (x)=n^{-1}{\sum }_{k=0}^{n-1}\phi (f^{k}x)$. Choose a point whose forward orbit has the prescribed convergence against every continuous function; the hypothesis gives a set of such points of full measure. For each continuous function $\phi$, telescoping givesbecause a continuous function on a compact metric space is bounded. Both test functions are continuous, so their averages converge to their prescribed integrals. Hencefor every continuous function $\phi$. The continuous-test criterion for an invariant measure therefore gives $f_{\ast }\mu =\mu$. The points in the hypothesis can now be called generic points for an invariant measure.

For ergodicity, let $B$ be a Borel set invariant modulo null sets, and set $h=1_B$, $p=\mu (B)$. Then $h\circ f=h$ almost everywhere. Since invariance preserves null preimages, deleting the union of all iterated preimages of the exceptional null set gives $h(f^{k}x)=h(x)$ for every $k\ge 0$ on one full-measure set. Thus $A_{n}h=h$ almost everywhere for every $n$.

Given $ϵ>0$, the continuous approximation of Borel indicators on a compact metric space, proved in the supporting steps below, supplies $\phi \in C(X)$ with $0\le \phi \le 1$ and $\Vert h-\phi {\Vert }_{L^1(\mu )}<ϵ$. By the hypothesis, $A_n\phi \to \int \phi d\mu$ almost everywhere. Since $0\le A_n\phi \le 1$, the dominated convergence theorem also gives convergence in $L^1(\mu )$. Invariance and the triangle inequality give the uniform estimateConsequently,Letting $n\to \infty$ and then $ϵ\downarrow 0$ proves $\Vert h-p{\Vert }_1=0$. Explicitly,so $p\in \{0,1\}$. Every invariant Borel set has measure zero or one, which provesThis proof controls measurable-set averages through a uniform $L^1$ estimate; convergence only against continuous test functions is not silently assumed to include indicator functions.

The printed symbol $E$ is not defined. Take $E=L^1(\mathbb{R})$, or more generally an $L^1$ subspace with its induced $L^1$ norm; density must be in that norm. Given $g\in E$, choose a sequence $f_n\in F$ withBy the assumed convergence on $F$, each $T_r{f}_n\to T_0{f}_n$ holds outside a null set $N_n$. Their countable union $N={\bigcup }_n{N}_n$ is still null. Thus outside $N$ the convergence statements for every member of this single approximation sequence hold simultaneously. No assertion of simultaneous convergence for the entire uncountable dense class is needed.

Use the approximants just chosen. The weak type (1,1) bound givesThe measures of these exceptional sets are summable. Indeed, the measure of their union for $n\ge m$ is at most $C{\sum }_{n\ge m}{2}^{-n}\to 0$. Hence their limsup is null; this is the Borel-Cantelli first lemma. Outside it, $S(g-f_n)\to 0$.

Remove also the countable null set from part (b). At each remaining point and for every $n$, the estimate in part (a) yieldsLetting $n\to \infty$ makes the right side zero. ThereforeThis proves almost-everywhere convergence from weak-type domination. The argument uses countably many measurable exceptional sets and does not assume that a supremum over all real parameters is measurable.

For independent observations in a natural exponential family, the log likelihood isIn a minimal exponential family the observed Fisher information matrix is positive definite, so the log likelihood is strictly concave. Any finite solution of the likelihood equation $\nabla \kappa (\hat{\theta })=\bar{Y}$ is therefore the unique global maximum likelihood estimator. Existence is a separate issue: the observed sufficient-statistic mean must belong to the image of the mean parameter map $\nabla \kappa$.

For a full regular minimal natural exponential family this image is the interior of the convex support of an exponential family, $C=\overline{\mathrm{conv}(\mathrm{supp}(h\nu ))}$. The necessity follows from a supporting hyperplane argument. If a finite parameter had its mean on a boundary face, some nonzero $v$ would satisfy $v^{T}Y\le b$ almost surely and $\mathbb{E}(v^{T}Y)=b$. Equality of the expectation forces $v^{T}Y=b$ almost surely. Exponential tilting preserves the base-measure null sets, so this contradicts minimality.

For sufficiency, let $\bar{Y}$ be an interior point of $C$. Choose finitely many points $y_1,\dots ,y_m$ of the support whose convex hull contains a ball about $\bar{Y}$, say of radius $2\epsilon$. Such points exist because $\bar{Y}$ is interior to the closed convex hull; one may approximate the vertices of a small simplex surrounding it by finite convex combinations of support points. Take support neighbourhoods of radius at most $\epsilon$ with positive finite base-measure masses $a_j$. These masses can be taken finite by restricting to bounded neighbourhoods and using finiteness of the normalizing integral at any fixed interior parameter. For every vector $\theta$, at least one $j$ has ${\theta }^T(y_j-\bar{Y})\ge 2\epsilon \Vert \theta \Vert$. Integrating over that neighbourhood yieldsThus the negative log likelihood grows to infinity as $\Vert \theta \Vert \to \infty$. At a finite boundary point of the full open natural parameter space, the normalizing integral is infinite, since otherwise that point would itself belong to the full domain. Fatou's lemma makes $\kappa (\theta )$ diverge on approach to such a boundary. A minimizing sequence for $\kappa (\theta )-{\theta }^T\bar{Y}$ therefore stays in a bounded subset away from the boundary; it has an interior limit where the minimum is attained. Its gradient vanishes, giving the required likelihood equation. Strict convexity gives uniqueness.

Hence, under these full, regular and minimal conventions,This is maximum likelihood existence in a full regular natural exponential family. Regularity alone does not make every sample admissible: in a Bernoulli distribution an all-zero or all-one sample has its sample mean on the boundary of $C=[0,1]$, and the canonical maximum is approached only as $\theta \to -\infty$ or $+\infty$. The usual probability parameter then has a boundary maximum in an enlarged model, but no finite canonical maximizer. A regular restriction of the natural parameter space requires membership in its actual mean-map image, rather than merely in $\mathrm{int}C$. Without minimality, affine redundancies may prevent uniqueness of the canonical parameter even when the fitted probability distribution is unique.

Set $Y_n={\sup }_{m\ge n}|f_m|$. It is an extended-real measurable function and decreases to $Y={\mathrm{limsup}}_m|f_m|$. If $f_n\to 0$ almost surely, then $Y_n\to 0$ almost surely. For any $ϵ>0$, the decreasing sets $\{Y_n>ϵ\}$ have intersection contained, up to a null set, in the exceptional set for convergence. Continuity from above of the probability measure gives $P(Y_n>ϵ)\to 0$.

Conversely, if $Y_n\to 0$ in probability, then $P(Y>ϵ)\le P(Y_n>ϵ)\to 0$ for every positive $ϵ$. Taking rational $ϵ>0$ gives $Y=0$ almost surely, so $|f_n|\to 0$ almost surely. This proves both directions, even when some individual $Y_n$ take the value infinity.

A statistic is sufficient when the conditional distribution of the observation given it does not depend on the parameter. It is minimal sufficient when it is, up to null sets, a measurable function of every sufficient statistic. For a dominated family, the likelihood-ratio criterion identifies two observations precisely when their likelihood ratio is independent of the parameter.

The Rao–Blackwell theorem says that conditioning a square-integrable unbiased estimator on a sufficient statistic preserves unbiasedness and cannot increase variance. The Cramér–Rao bound for an unbiased estimator of scalar $\theta$ is ${\mathrm{Var}}_{\theta }\hat{\theta }\ge 1/I(\theta )$. It requires parameter-independent support, differentiability, interchange of differentiation and integration, and finite positive Fisher information. More generally the numerator is the square of the derivative of the estimand. The bound follows from $E_{\theta }[\hat{\theta }{\partial }_{\theta }\log f]=1$ and Cauchy–Schwarz.

For the uniform sample, the joint likelihood is ${\theta }^{-n}{1}_{\{0<x_i<\theta \forall i\}}$. Its dependence on the observations is through $T={\max }_i{X}_i$, proving sufficiency. Two such likelihoods have parameter-independent ratio exactly when their maxima coincide, proving minimality. The density of $T$ is $n{t}^{n-1}/{\theta }^n$ on $(0,\theta )$. Unbiasedness of $h(T)$ saysDifferentiating this absolutely continuous identity gives $h(t)=(n+1)t/n$ almost everywhere. Conversely that function is unbiased. Every finite-variance unbiased estimator, after Rao–Blackwellization, must therefore give this same function of $T$. Its variance cannot exceed that of any such estimator. ThusThe Cramér–Rao argument does not apply because the support depends on $\theta$; in particular differentiation of the support boundary invalidates the regular score identity.

An arbitrage is a portfolio with nonpositive initial cost, nonnegative terminal payoff almost surely and a positive payoff with positive probability, allowing unused initial cash to be invested in the riskless asset. Equivalently it is a zero-cost portfolio with those terminal properties. An equivalent martingale measure $Q$ has the same null sets as the physical measure and satisfies $E_Q[S_1^i]=(1+r)S_0^i$ for every traded asset, with appropriate integrability.

For an agent with differentiable increasing concave utility and an optimal portfolio, let $W^{\ast }$ be its terminal wealth. The first-order conditions for variations in risky holdings are $E[U^{\prime }(W^{\ast })(S_1^i-(1+r)S_0^i)]=0$. If $U^{\prime }(W^{\ast })>0$ and integrable, normalize it to the probability density $Z=U^{\prime }(W^{\ast })/E[U^{\prime }(W^{\ast })]$. The first-order conditions make $Q$ with $dQ=ZdP$ an equivalent martingale measure. A small additional payoff $H$ then has marginal priceThis follows by setting the first variation of expected utility for buying a small claim at price $p$ equal to zero.

We construct the density and singular part, rather than assuming the desired Lebesgue decomposition theorem. Put $\rho =\mu +\nu$, a finite dominating measure, and use the real Hilbert space $L^2(\rho )$. The linear functionalis well defined on its equivalence classes, since a $\rho$-null set is $\nu$-null. The Cauchy-Schwarz inequality and $\nu \le \rho$ giveBy the Riesz representation theorem, there is $h\in L^2(\rho )$ with $T(g)=\int ghd\rho$. Indicators of all measurable sets belong to this Hilbert space, soThe nonnegativity of these two measures forces $0\le h\le 1$ almost everywhere. For instance, if $\rho (\{h\le -1/n\})>0$, its $\nu$-measure would be negative; the analogous sets $\{h\ge 1+1/n\}$ contradict positivity of $\mu$. Their countable union excludes every violation. Modify $h$ on its measurable $\rho$-null exceptional set so that these inequalities hold everywhere. This is the Hilbert-space construction of dominated measure densities.

DefineThen $B\in \Sigma$, $f$ is a nonnegative measurable function, and $\mu (B)={\int }_B(1-h)d\rho =0$. The identity for integration against $\mu$ follows first for indicators, then for nonnegative simple functions and finally for all nonnegative functions by the monotone convergence theorem. Applying it to $f{1}_A$ givesIn particular ${\int }_{\Omega }fd\mu =\nu (\Omega \setminus B)<\infty$, so $f\in L^1(\mu )$. Additivity on the disjoint measurable sets $A\setminus B$ and $A\cap B$ now givesThis is the Lebesgue decomposition from a sum-measure density: the first measure has a density, and the second is concentrated on a $\mu$-null set, so is mutually singular with $\mu$.

The absolute continuity of measures condition $\nu \ll \mu$ means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If this holds, then $\nu (B)=0$, eliminating the singular part above. Conversely, a measure of the form $\nu (A)={\int }_{A}fd\mu$ vanishes on every $\mu$-null set. ThereforeThus absolute continuity is exactly the absence of the singular part in this construction.

Construct the densities using the Riesz representation theorem for a Hilbert space, rather than assuming the measure-decomposition conclusion. Put $\rho =\mu +\nu$, a finite dominating measure, and work in the real L2 space $L^2(\rho )$. The linear functionalis well-defined on its equivalence classes because $\nu \le \rho$. By the Cauchy-Schwarz inequality,Thus the Riesz representation theorem supplies $h\in L^2(\rho )$ such that $L(g)=\int ghd\rho$. Since $\rho$ is finite, $h$ is also integrable. Testing $g=1_A$ givesThe first identity and nonnegativity of $\nu$ rule out a positive-measure set where $h<0$; for example, test $\{h\le -1/n\}$. The second identity and nonnegativity of $\mu$ similarly rule out $\{h\ge 1+1/n\}$. Modify a measurable representative on a $\rho$-null set so that $0\le h\le 1$ everywhere. This is the Hilbert-space construction of dominated measure densities.

SetThen $B\in \Sigma$ and $\mu (B)=0$. The indicator identities for $\mu$ extend to integrals of nonnegative measurable functions by approximation with simple functions and the monotone convergence theorem. Consequently,and, for every $A$,HenceThis proves Lebesgue decomposition from a sum-measure density. The argument includes the cases of zero measures; there is no division on $B$, where the denominator vanishes.

Absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If $\nu \ll \mu$, then $\nu (B)=0$ in the construction above, so the singular term vanishes. Conversely, if $\nu (A)={\int }_{A}fd\mu$ with a nonnegative integrable $f$, that integral is zero on every $\mu$-null set. ThusThis is the finite-measure Radon-Nikodym theorem as a consequence of the proved construction.

Finally restrict the base measure to the sub-sigma-algebra ${\Sigma }_0$. For a real $f\in L^1(\mu )$, write $f=f^{+}-f^{-}$, with both parts nonnegative and integrable. On ${\Sigma }_0$ define finite positive measuresThey are absolutely continuous with respect to $\mu {|}_{{\Sigma }_0}$. The result just proved gives nonnegative, ${\Sigma }_0$-measurable densities $g_{\pm }$ satisfyingSet $f_0=g_{+}-g_{-}$. Since $\int (g_{+}+g_{-})d\mu =\int |f|d\mu <\infty$, this is in $L^1(\Omega ,{\Sigma }_0,\mu )$, and subtraction givesFor complex $f$, apply the real construction separately to its real and imaginary parts and combine the two results. This is conditional expectation from finite-measure densities. The resulting conditional expectation is unique almost everywhere: for two real versions, their difference has integral zero on its positive and negative level sets in ${\Sigma }_0$, forcing it to vanish; apply this to both components in the complex case. No probability normalization of $\mu$ is required.