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On a finite measure space, if measurable functions are uniformly bounded and converge pointwise, then their integrals converge. This is the dominated convergence theorem with a constant dominating function.

For every finite measure $\mu$ on the predictable sigma-algebra, bounded simple predictable processes are dense in $L^2(\mu )$. Include a bounded ${\mathcal{F}}_0$-measurable value supported at time zero when $\mu$ may charge that slice. Indicators of the generating rectangles and the Pi-lambda theorem prove density. This gives the completion step in the Itô isometry construction.

The angular-frequency Fourier transform of a finite measure, or a complex measure of finite total variation, is defined by the displayed integral. It is bounded by total variation and continuous by dominated convergence. An integrable density recovers the ordinary Fourier transform of a function. A measure supported on a curved surface instead leads to a Fourier extension operator.

On a finite measure space, the convergence in the Birkhoff ergodic theorem also holds in $L^1$. Prove it first for bounded truncations by the dominated convergence theorem, then use the $L^1$ contraction $\Vert A_{n}h{\Vert }_1\le \Vert h{\Vert }_1$ and Fatou lemma to remove the truncation.

If a finite measure on ${\mathbb{R}}^n$ has square-integrable Fourier transform, the Plancherel theorem gives an $L^2$ inverse transform $g$. Pairing with every Schwartz function and using Fourier inversion identifies the measure with $gdx$. This establishes absolute continuity, rather than presuming it. For a positive measure, $g\ge 0$ and $g\in L^1$ as well. Multiplication by $f\in L^{\infty }(\mu )$ yields $\Vert \widehat{fd\mu }{\Vert }_2\le \Vert \hat{\mu }{\Vert }_2\Vert f{\Vert }_{L^{\infty }(\mu )}$.

Let $X\ge 3$, and let $f$ be a normalized multiplicative arithmetic function supported on squarefree integers, with $|f(p)|\le 1$. Form the finite measure $\nu ={\sum }_{n}f(n)n^{-1-1/\log X}{\delta }_{\log n}$ using Dirac measures. Under the transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$, its Euler product is ${\prod }_p(1+f(p)p^{-1-1/\log X-it})$. Expanding its logarithmic modulus, and replacing the damped prime weights by $1/p$ up to $X$ with Mertens first theorem, gives the displayed absolute bound, uniformly in real $t$. Hence $|\hat{\nu }(t)|\ge \delta \log X$ forces $D(f,n^{it};X{)}^2\le \log (C/\delta )$. The distance expression still defines a nonnegative pretentious score for prime values bounded by one; its actual metric interpretation uses unit-modulus values.

Both functions represent the same $L^1$ equivalence class. Fubini's theorem shows that the interval-mass function has derivative $Du$, so its difference from $u$ is constant. Continuity of a finite measure gives left continuity for $u_l$ and right continuity for $u_r$. Their opposite one-sided limits differ by the measure atom $Du(\{t\})$. There are only countably many nonzero measure atoms, since every set of measure atoms above a fixed magnitude threshold is finite.

For the unit-modulus prime values in this question, the Granville-Soundararajan distance isThe second identity uses $|f(p)|=|g(p)|=1$. Thus it is the Euclidean distance between the finite weighted vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$ and $(g(p)/\sqrt{2p}{)}_{p\le X}$. Nonnegativity, symmetry and the triangle inequality follow from that norm representation, and it vanishes exactly when the two prime-value vectors agree. This proves the prime-restriction metric for pretentious distance.

On whole arithmetic functions at fixed $X$ it is a pseudometric. For example $f=1$ and $g={\mu }^2$ have identical values at every prime and zero distance for every $X$, but different values at four. Thus the literal identity-of-indiscernibles assertion needs the prime-restriction quotient.

Since $\mu (p)=-1$,For completeness, partial summation and the prime-counting form of the Prime number theorem giveTo verify the constant and the error, replace $\pi$ by $\mathrm{Li}+E$ in the first expression. Differentiating $\mathrm{Li}(u)/u$ shows its two contributions combine to $\log \log X$ plus a constant. The error integral converges absolutely because $|E(u)|/u^2\ll e^{-c\sqrt{\log u}}/u$, and $E(X)/X\to 0$. This proves the Mertens second theorem in the form needed here. In particular the pretentious distance between one and the Möbius function satisfies

For the last request, retain the earlier prime normalization $|f(p)|=1$; the proof also works with the weaker assumption $|f(p)|\le 1$. Without a bound on prime values, the original definition and the absolute convergence of the displayed measure need not apply to an unrestricted multiplicative arithmetic function. For example $f(n)=\mu (n{)}^{2}n$ is multiplicative and squarefree-supported, but for $X>e$ the displayed measure has infinite positive mass: its prime terms alone are ${\sum }_p{p}^{-1/\log X}$, which diverges by the Prime number theorem.

Put $L=\log X$, $\sigma =1+1/L$, and use the Fourier transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$. Since $f$ is supported on squarefree integers, the defining property of a multiplicative arithmetic function gives the absolutely convergent Euler productThe bound on prime values gives $|f(n)|\le 1$ for squarefree $n$, so the total variation norm of a measure is bounded by $\zeta (\sigma )$; thus $\nu$ is a finite measure. Expanding the logarithm of the modulus, with an absolute $O(1)$ remainder since ${\sum }_p{p}^{-2\sigma }\ll 1$, givesAll constants here are uniform in $t$. Moreoverusing the pole of the Riemann zeta function. The nonnegative deficits $1-\mathrm{Re}(f(p)p^{-it})$ giveIndeed, the replacement of $p^{-\sigma }$ by $1/p$ for $p\le X$ costs at mostby the Mertens first theorem, while discarding the $p>X$ terms costs nothing in this lower bound. Thus the large Fourier coefficient criterion for pretentiousness isIn particular,The comparison function is the Archimedean character $n^{it}$ with this sign because our Fourier transform uses $e^{-itu}$.

Use the Fourier transform convention $\hat{\nu }(\xi )=\int e^{-ix\cdot \xi }d\nu (x)$ throughout this question. The decay assumption implies square integrability, since polar coordinates giveNear zero the integrand is bounded by $r$, and at infinity it is bounded by $r^{-1-2ϵ}$. The positive $ϵ$ is what makes the latter integrable.

By the Plancherel theorem, there is a function $g\in L^2({\mathbb{R}}^2)$ whose Fourier transform is $\hat{\mu }$. It is the density of $\mu$ with respect to Lebesgue measure. To justify this step rather than assume a density, for every Schwartz function $\phi$, Fourier inversion givesThe finite measure and the locally integrable function thus define the same tempered distribution, so they agree as measures: $d\mu =gdx$. In particular $g\ge 0$ almost everywhere and $\int g=\mu ({\mathbb{R}}^2)<\infty$. This is the L2 density from a square-integrable Fourier transform principle.

For $f\in L^{\infty }(\mu )$, the density of $fd\mu$ is $fg$. The inequality $|f|\le \Vert f{\Vert }_{L^{\infty }(\mu )}$ holds wherever $g>0$, apart from a Lebesgue measure zero set, soA second application of the Plancherel theorem yields the explicit boundThe implicit constant in the requested estimate may depend on the measure and its Fourier-decay bound, but is independent of $f$.

The structure theorem for functions of bounded variation is the measure representation of the distributional derivative: for $u\in BV(\Omega )$ there is a unique finite vector Radon measure $Du$ withfor every $\phi \in C_c^1(\Omega ;{\mathbb{R}}^n)$. It also has a polar decomposition of a vector measure $Du={\sigma }_u|Du|$ with $|{\sigma }_u|=1$ for $|Du|$-almost every point.

To prove it, let $L(\phi )=-\int u\mathrm{div}\phi$. The test-function definition of total variation seminorm, applied to both signs, gives $|L(\phi )|\le \mathrm{TV}(u)\Vert \phi {\Vert }_{\infty }$. Compactly supported smooth vector fields are uniformly dense in $C_0(\Omega ;{\mathbb{R}}^n)$, so $L$ extends uniquely to a bounded functional there. The Riesz-Markov-Kakutani representation theorem, applied componentwise, gives the unique vector measure $Du$. The operator norm of $L$ is exactly the defining variation supremum; the vector-measure dual norm is $|Du|(\Omega )$, proving equality. Conversely any finite measure satisfying the identity bounds that supremum, so this also characterizes membership in the BV space.

The Radon-Nikodym theorem applied to the components relative to $|Du|$ gives ${\sigma }_u$; the definition of the variation measure forces $|{\sigma }_u|=1$ almost everywhere. If $u$ is smooth, ordinary integration by parts gives $Du=\nabla u{\mathcal{L}}^n$. Compact support of the test field removes any boundary contribution; no boundary regularity is needed for this representation statement.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

Use a filtered probability space $(\Omega ,\mathcal{F},({\mathcal{F}}_t),\mathbb{P})$, and put ${\mathcal{F}}_{0-}={\mathcal{F}}_0$. The predictable sigma-algebra on $\Omega \times [0,\infty )$ isEquivalently it is generated by left-continuous adapted processes. A real-valued previsible process is a function $(\omega ,t)\mapsto H_t(\omega )$ that is measurable for $\mathcal{P}$ and the Borel sigma-algebra of $\mathbb{R}$.

A simple predictable process has the formwhere $h_{-1}$ is bounded and ${\mathcal{F}}_0$-measurable, and each $h_i$ is bounded and ${\mathcal{F}}_{t_i}$-measurable. Set $H=0$ after $t_n$; allowing a final infinite interval gives an equivalent dense class. The values may be restricted to finitely many values if desired, by approximating bounded measurable functions by simple functions. The initial-time term matters for arbitrary finite measures on $\mathcal{P}$: a measure can charge time zero, even though the stochastic integral against a continuous martingale does not.

Let $V$ be the closure of the simple predictable processes in the Hilbert space $L^2(\mathcal{P},\mu )$. Common refinement of deterministic partitions makes these processes a vector space. To prove density of simple predictable processes for finite measures, suppose $g\in V^{\perp }$. Since $\mu$ is a finite measure, the Cauchy-Schwarz inequality gives $g\in L^1(\mu )$, sodefines a finite signed measure. The indicator of each predictable generating rectangle, including each time-zero rectangle, is a simple predictable process; hence $\nu$ vanishes on these rectangles.

Finite intersections of the rectangles are again rectangles or empty. Add the whole space to this pi-system. Its $\nu$-mass is zero because $1_{\Omega \times [0,n]}$ is simple and converges to $1$ in $L^2(\mu )$ by the dominated convergence theorem. The zero sets of $\nu$ form a lambda-system now that its total mass is zero. The Pi-lambda theorem therefore gives $\nu (E)=0$ for every $E\in \mathcal{P}$. Taking the sets where $g>0$ and $g<0$ gives $g=0$ $\mu$-almost everywhere.

Thus $V^{\perp }=\{0\}$ and the orthogonal complement criterion for a closed subspace of a Hilbert space yieldsOmitting the time-zero term would make this conclusion false when $\mu$ is concentrated at time zero.

The Birkhoff ergodic theorem states: for a measure-preserving transformation $\theta$ of a finite measure space and $f\in L^1$, the averages $A_{n}f=n^{-1}{\sum }_{j=0}^{n-1}f\circ {\theta }^j$ converge almost everywhere and in $L^1$ to an invariant function $f^{\ast }$. On a probability space,where $\mathcal{I}$ is the invariant sigma-algebra modulo null sets. In particular the integral is preserved, and if $\theta$ is ergodic then $f^{\ast }=\int fd\mu$ almost everywhere; for a finite space of nonunit mass, this constant is $\mu (E{)}^{-1}\int fd\mu$ when $\mu (E)>0$.

The maximal ergodic lemma says that for real integrable $f$, with $S_{k}f={\sum }_{j=0}^{k-1}f\circ {\theta }^j$ and $E_N=\{{\max }_{1\le k\le N}{S}_{k}f>0\}$,The same inequality holds on $E_{\infty }=\{{\sup }_{k\ge 1}{S}_{k}f>0\}$, by taking the increasing limit and dominated convergence. These statements do not require the transformation to be invertible or the space to be ergodic.

TakeThis is a finite measure, hence a sigma-finite measure. If $\nu (B)=0$, nonnegativity gives $\mu (B)={\mu }_0(B)=0$, so both $\mu$ and ${\mu }_0$ are absolutely continuous with respect to $\nu$.