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For a finite measure on a measurable space whose singletons are measurable sets, the set of point atoms of a measure is at most countable. For each positive integer $m$, let $D_m$ consist of the points with singleton measure at least $1/m$. If $D_m$ contained more than $m\mu (\Omega )$ distinct points, finite additivity on those disjoint singletons would exceed the total measure. Thus each $D_m$ is finite.

Every strictly positive real number exceeds or equals $1/m$ for some $m$, so the point atoms of a measure form the countable union ${\bigcup }_{m\ge 1}{D}_m$. This proves that the set is countable. It also shows that a cumulative distribution function has only countably many jumps. The result does not say the measure is concentrated on those points: a non-atomic measure may assign them total measure zero.

If $E\subseteq [0,1]$ is a Lebesgue measurable set and $0\le b\le \lambda (E)$, some measurable subset $B\subseteq E$ has Lebesgue measure $b$. The function $F(t)=\lambda (E\cap [0,t])$ has Lipschitz continuity with constant $1$, starts at $0$ and ends at $\lambda (E)$; the intermediate value theorem proves the assertion. Applying the assertion successively to remaining portions of $E$ gives pairwise disjoint measurable pieces with any finite list of nonnegative measures whose sum is at most $\lambda (E)$. This is a concrete consequence of non-atomic measure structure, which fails for general measures containing an atom of a measure.

The answer isIndeed, Lebesgue measure on $E\subset {\mathbb{R}}^n$ is a sigma-finite measure, and duality of Lp spaces gives $(L^1(E){)}^{\ast }=L^{\infty }(E)$. The restricted Lebesgue measure is a non-atomic measure and $E$ has positive measure, so the Dual of L infinity is larger than L1: the canonical integral representation does not exhaust $(L^{\infty }(E){)}^{\ast }$.

For $1<p<\infty$, the Holder conjugate exponent also lies strictly between one and infinity. Duality of Lp spaces then holds in both directions, in agreement with Reflexivity of Lp spaces. For $p=\infty$, the first required identification already fails.

Equip $S$ with a sigma-algebra $\mathcal{S}$. A Poisson random measure with intensity measure of a point process $\mu$ is a random counting measure $\Pi$ such that, for every measurable $A$ with $\mu (A)<\infty$,and the counts on every finite collection of pairwise disjoint measurable sets are independent. Thus $\mathbb{E}\Pi (A)=\mu (A)$. For a sigma-finite measure $\mu$, sets of infinite intensity contain infinitely many points almost surely: exhaust such a set by increasing finite-intensity sets and use the Poisson count tails as their means tend to infinity. The count is a measurable random variable for every measurable $A$, which is part of saying that $\Pi$ is a random measure.

Under the usual point representation write $\Pi ={\sum }_j{\delta }_{Z_j}$, counting multiplicities. If $\mu$ is a non-atomic measure, this is a simple Poisson point process; if $\mu$ has atoms, several point occurrences can have the same location, and each occurrence is counted separately. In a topological setting, local finiteness requires that the intensity be finite on compact sets.

Sufficient existence conditions are that $S$ is a standard Borel space and $\mu$ is a sigma-finite measure. This gives existence of a Poisson random measure on a standard Borel space. If a simple locally finite configuration is desired, additionally take a non-atomic, locally finite intensity on a locally compact second-countable Hausdorff space. These conditions are sufficient, not claimed necessary. In particular sigma-finiteness suffices for a countable family of occurrences that can each be independently coloured.

Let $R$ and $G$ be the red and green counting measures. The marking assumption means that, conditional on $\Pi$, the colours of all point occurrences are independent, with the same red probability $r$. For a measurable set $A$ of finite intensity $m=\mu (A)$, condition on its total count. To obtain $a$ red and $b$ green points the total must be $a+b$, so direct multiplication gives the joint red and green Poisson count formulaThus the two counts are independent Poisson variables with respective means $rm$ and $(1-r)m$. This computation includes zero intensity with the usual convention for the probability of zero counts.

For disjoint finite-intensity sets $A_1,\dots ,A_k$, the original counts are independent. Conditional marking in different sets uses disjoint collections of independent colours. Repeating the preceding calculation therefore givesThis proves the disjoint-count definition for each coloured Poisson point process and independence of their two count vectors.

To obtain independence of the whole processes, take any finite list of red test sets and any finite list of green test sets, initially all of finite intensity. Refine their union into the disjoint cells given by their membership patterns. Every test-set count is the sum of the counts of its cells. The entire red cell vector is independent of the entire green cell vector by the product formula, hence so are the corresponding test-set count vectors. For arbitrary measurable test sets, intersect first with a finite-intensity exhaustion of $S$ and pass to the increasing count limits. The count cylinder events generate the two evaluation sigma-algebras, and independence extended from generating pi-systems gives independence of those sigma-algebras. ConsequentlyThis is a direct count proof of the Poisson thinning theorem; no general marking or thinning result has been used to deduce it.

All vector spaces in this solution are complex, while convex combinations use real coefficients. An extreme point $x$ of a convex set $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.

The Krein-Milman theorem says that every nonempty compact convex set $K$ in a Hausdorff locally convex space is the closed convex hull of its extreme points. Here is a proof. A face of a convex set $F\subseteq K$ is a convex set such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contains distinct $x,y$, a continuous real linear functional $\ell$ on the underlying real locally convex space distinguishes them. Such a linear functional exists because the space is Hausdorff and locally convex, by the Hahn-Banach theorem. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an extreme point. The same argument inside any nonempty compact face of $K$ supplies an extreme point of $K$ lying in that face.

Let $D$ be the closed convex hull of the extreme points of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the Hahn-Banach separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an extreme point $e$ of $K$. But $e\in D$ and $\ell (e)\ge \ell (x_0)>{\sup }_D\ell$, a contradiction. Thus $\boxed{K=\overline{\mathrm{co}}(\mathrm{ext}K)}$, proving Krein-Milman theorem.

For a nonempty compact Hausdorff space $K$, the extreme points of the dual unit ball of C(K) areThis is the permitted description without proof, with $C(K)$ denoting the complex space of continuous functions on a compact space equipped with the supremum norm. If $K$ is empty, $C(K)=\{0\}$ and the closed unit ball of its continuous dual space has the single extreme point $0$ instead.

The complex Banach–Stone theorem states that a surjective complex-linear isometric isomorphism of normed spaces $T:C(K)\to C(L)$ has the formwhere $u\in C(L)$ and $\phi :L\to K$ is a homeomorphism. Conversely, every such map is a surjective complex-linear isometric isomorphism of normed spaces for the supremum norm.

To prove this, suppose first that $K,L$ are nonempty. The Banach-space adjoint $T^{\ast }$ is a bijective isometric isomorphism of normed spaces on the continuous dual spaces, so it bijects their closed unit balls and preserves extreme points. The displayed extreme point description gives uniquelyUniqueness follows by evaluating at the constant function $1$, and then using that continuous functions separate points of a compact Hausdorff space. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^{\ast }$ on extreme points proves surjectivity of $\phi$: the preimage of any ${\delta }_x$ is $\alpha {\delta }_y$ for some $y$. If $\phi (y_1)=\phi (y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\phi$ is bijective.

For every $f\in C(K)$, $f\circ \phi =(Tf)/u$ is continuous. The evaluation map $K\to {\mathbb{C}}^{C(K)}$, $x\mapsto (f(x){)}_f$, is a continuous injection of a compact Hausdorff space into a Hausdorff product topology, hence a homeomorphism onto its image. Continuity of every coordinate $f\circ \phi$ proves continuity of $\phi$. A continuous bijection between compact Hausdorff spaces is a homeomorphism. Conversely the weighted-composition formula plainly preserves the supremum norm, and its inverse isIf one compact space is empty, a surjective isometric isomorphism of normed spaces forces the other to be empty, and the empty homeomorphism gives the corresponding trivial case. This completes Banach–Stone theorem.

Neither the space of sequences converging to zero $c_0$ nor $L^1[0,1]$ can be isometrically a continuous dual space of a Banach space. Indeed, every nonzero continuous dual space has a nonempty weak-star compact closed unit ball by Banach-Alaoglu theorem, and Krein-Milman theorem then guarantees an extreme point. A bijective linear isometry preserves extreme points of closed unit balls.

The closed unit ball of $c_0$ has no extreme points. Given $x\in c_0$ with $\Vert x{\Vert }_{\infty }\le 1$, choose $n$ with $|x_n|<1$ and $0<\epsilon <1-|x_n|$. The two distinct elements $x\pm \epsilon e_n$ remain in that closed unit ball and have midpoint $x$.

The closed unit ball of the complex Lp space $L^1[0,1]$ also has no extreme points. An element of norm less than one can be perturbed by a sufficiently small nonzero Lp space element in both directions. If $\Vert f{\Vert }_1=1$, the non-atomic measure $|f(t)|dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto {\int }_0^s|f(t)|dt$ attains $1/2$. Put $h=f(1_A-1_{A^c})$. For $0<\epsilon <1$, the distinct functions $f\pm \epsilon h$ both have Lp normwith the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus neither proposed space is isometrically a Banach dual.

Finally, $[0,1]$ is a connected space, whereas $[0,1]\cup [2,3]$ is disconnected. They cannot be homeomorphic. By Banach–Stone theorem, $C[0,1]$ and $C([0,1]\cup [2,3])$ are not isometrically isomorphic as complex Banach spaces.

Consider the Borel setIts indicator function is nonnegative and measurable, so the Tonelli theorem applies to the product measure, without needing any separate absolute-integrability argument. At a fixed $y\in (a,b]$, its section in the first coordinate is $(a,y]$; outside $(a,b]$ that section is empty. ThereforeThe last equality uses the definition of a cumulative distribution function on the half-closed interval. It retains all point atoms of a measure on the diagonal $x=y$, excludes a possible point atom of a measure at $x=a$, and includes $y=b$, exactly as required by the set $D$. If $a=b$, both sides vanish. This verifies the identity even when the probability measures are not non-atomic measures.^[ref]