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For $t\in \mathbb{R}$, define the set functionalThe ROF level-set formulation states that the unique minimizer of $J(u)=\Vert u-f{\Vert }_2^2/2+\alpha \mathrm{TV}(u)$ has superlevel sets minimizing $F_t$ for almost every $t$. Conversely, an admissible $u$ with minimizing superlevel sets is the ROF minimizer. Thus the concise characterization is

The signed layer-cake identity for quadratic fidelity and the coarea formula for BV functions giveIndeed $u^2/2-fu={\int }_{\mathbb{R}}(t-f)({\chi }_{\{u>t\}}-{\chi }_{\{0>t\}})dt$ pointwise. Its absolute integral is bounded by $u^2/2+|fu|$, so Fubini's theorem applies for $u,f\in L^2$. The negative-level baseline is essential; integrating the unadjusted $F_t$ would generally diverge.

Each $F_t$ attains its minimum. A minimizing sequence of indicator functions has bounded $L^1$ norm, and comparison with the empty set bounds its perimeter by an $L^1$ forcing bound. Bounded-variation compactness supplies a limiting indicator function. The forcing integral converges by dominated convergence, while relative perimeter is sequentially lower semicontinuous.

For $s<t$, the supplied comparison lemma applied to $(f-t)/\alpha <(f-s)/\alpha$ makes every selected minimizer $A_t$ contained in $A_s$ up to a null set. The comparison of perimeter minimizers with ordered forcing also follows directly: compare $A_t$ with $A_t\cap A_s$, compare $A_s$ with $A_t\cup A_s$, add and use submodularity of relative perimeter to obtain $(t-s)|A_t\setminus A_s|\le 0$.

Select minimizers at rational levels, remove their countably many exceptional null sets, and reconstruct $v(x)=\sup \{q\in \mathbb{Q}:x\in A_q\}$. The strict superlevel set is $\{v>t\}={\bigcup }_{q>t}{A}_q$. This union also minimizes $F_t$: take $q\downarrow t$, use monotone $L^1$ convergence of its indicator functions and sequential lower semicontinuity, and note that $m(t)={\min }_E{F}_t(E)$ satisfies $|m(t)-m(s)|\le |\Omega ||t-s|$.

To justify finite energy before assuming it, clip this reconstruction to $v_M\in [-M,M]$. Comparison with the empty set gives $\alpha \mathrm{Per}(E_t;\Omega )\le {\int }_{\Omega }|t-f|$ uniformly on a bounded interval of levels. Since $v_M=-M+{\int }_{-M}^M{\chi }_{E_t}dt$, pairing with compactly supported test-field divergences bounds its total variation seminorm by ${\int }_{-M}^M\mathrm{Per}(E_t;\Omega )dt<\infty$. Thus $v_M$ belongs to the BV space before applying the coarea formula for BV functions. The layer-cake argument on the finite interval $[-M,M]$ shows $J(v_M)\le J(T_{M}w)$ for every finite-energy competitor $w$, where $T_M$ is clipping. In particular $w=0$ bounds the $L^2$ norms and variations uniformly. Fatou's lemma excludes infinite values of $v$ on a positive-measure set. Bounded-variation compactness and weak convergence in $L^2$ identify an admissible limit $v$. For any fixed competitor, $T_{M}w\to w$ in $L^2$ and its variation tends to that of $w$, by bounded-variation contraction under clipping and sequential lower semicontinuity. Thus $J(v)\le J(w)$.

The quadratic fidelity is strictly convex, so every ROF minimizer equals $v$ almost everywhere and has the selected minimizing superlevel sets. Conversely, for any $u$ whose levels minimize $F_t$, integrate the levelwise inequality against any competitor's levels in the displayed identity to obtain $J(u)\le J(w)$. This proves both directions for signed as well as nonnegative data.

Put $\mu =Du$, a finite signed Radon measure, and defineFor $\phi \in C_c^1((a,b))$, Fubini's theorem for the signed measure givesThus $D{F}_l=\mu =Du$. A distribution on a connected interval with zero derivative is constant: any compactly supported test of integral zero is the derivative of a compactly supported test, so it pairs to zero with $u-F_l$. Choose the resulting constant $c$. Thenare one-sided representatives of a one-dimensional BV function. The first equals $u$ almost everywhere. Their difference is $\mu (\{t\})$, nonzero at at most countably many points, so the second also equals $u$ almost everywhere.

Finite-measure continuity applied to $|\mu |$ proves $F_l(s)\to F_l(t)$ as $s\uparrow t$, and $F_r(s)\to F_r(t)$ as $s\downarrow t$. Moreover the opposite one-sided limits are $F_l(t+)=F_r(t)$ and $F_r(t-)=F_l(t)$. Consequently both representatives are continuous exactly where $\mu (\{t\})=0$. For each positive integer $m$, there are only finitely many measure atoms of magnitude at least $1/m$, since their total magnitudes are bounded by $|\mu |((a,b))$. Their union is countable, proving the requested discontinuity bound. This does not require the jump points to be isolated; they can be dense.

The hint's one-dimensional statement is consistent with this construction: zero ${\mathcal{H}}^0$ measure means an empty set, so $S_u\subseteq J_u$. This concerns approximate discontinuities of the BV space class; arbitrary changes of representative at points could create artificial pointwise discontinuities. The interval-mass representatives above remove that ambiguity and give the required one-sided continuity directly.

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.

Use probability measure-preserving systems: $T:X\to X$ is measurable and $\mu (T^{-1}A)=\mu (A)$. Its invariant sigma-algebra isErgodicity means that every invariant measurable set has measure zero or one:Equivalently, every invariant measurable function is constant almost everywhere. Indeed all rational sublevel sets of a real invariant function belong to $\mathcal{I}$, so the zero-or-one alternative forces its distribution to concentrate at one value. Apply this to real and imaginary parts for complex functions. Conversely, an invariant set has an invariant indicator, so the function criterion implies the set criterion. This is the invariant-function characterization of ergodicity.

For an irrational rotation of the circle, $T(x)=x+\alpha (\mathrm{mod}1)$ with $\alpha \notin \mathbb{Q}$, an invariant $g\in L^2(m)$ has Fourier coefficients satisfyingFor $k\ne 0$ the multiplier is nonzero. Completeness of the Fourier series basis therefore makes $g$ constant. Apply this to an invariant indicator to prove the irrational rotation is ergodic. A second example is the cyclic permutation of $q$ points with equal masses: a nonempty invariant set contains the whole cycle and hence has measure one.

The identity on $[0,1]$ with Lebesgue measure is not ergodic, since $[0,1/2]$ is invariant and has measure $1/2$. Nor is rotation by $1/5$ on the circle: the setis invariant and has measure $1/2$. These examples also show why preservation of measure does not imply ergodicity.

Let $U=U_T$ be the Koopman operator, $Uf=f\circ T$, and setThe Von Neumann mean ergodic theorem says that, for $f\in L^2(\mu )$, these Cesaro averages converge in $L^2$ to the orthogonal projection onto the invariant functions:For an invertible system, $U$ is a unitary operator. The Hilbert-space theorem states more generally that the Cesaro averages of any unitary operator converge strongly to its fixed-space projection.

Here is the requested invertible-case proof. Since $U^{\ast }=U^{-1}$,Thus the orthogonal decomposition for unitary ergodic averages isOn the first summand, $A_{N}f=f$. On a vector $(I-U)g$, telescoping givesBecause $\Vert A_N\Vert \le 1$, approximation extends this convergence to the closure of the range. The orthogonal decomposition now proves the theorem for every $f$. An invariant $L^2$ function is $\mathcal{I}$-measurable, and an $\mathcal{I}$-measurable function is invariant by approximation with indicators; hence the projection is the stated conditional expectation.

In an ergodic system, invariant functions are constants by the criterion already proved. Their orthogonal projection is obtained by taking the inner product with $1$, whose norm is one. Therefore the ergodic limit is the space average:

The Birkhoff ergodic theorem states that, for $f\in L^1(\mu )$,The limit is invariant, integrable, and has the same integral as $f$. On a probability space convergence also holds in $L^1$. In an ergodic system the limit is $\int fd\mu$ almost everywhere.

For the topological sharpening, take a continuous map $T$ on a compact metric space. Unique ergodicity means there is exactly one invariant Borel probability measure $\mu$. The uniform ergodic convergence for uniquely ergodic systems says that every continuous real or complex $f$ satisfiesThus convergence holds at every point and uniformly in that point. The continuity, compactness and continuous-observable hypotheses belong to this theorem; it is not a claim of uniform convergence for all measurable $L^1$ functions.

Use a standard Borel probability space so that conditional measures can be realized by measures. The disintegration theorem for a probability measure states the following. For a measurable map $\pi :X\to Y$ between standard Borel spaces and $\nu ={\pi }_{\ast }\mu$, there is a measurable probability kernel $y\mapsto {\mu }_y$, unique for $\nu$-almost every $y$, such thatMeasurability means $y\mapsto {\mu }_y(A)$ is measurable for every Borel set $A$. More generallyfor integrable $f$, with equality of the last expression almost everywhere. The ${\mu }_y$ are the conditional measures of a factor. This also defines conditional measures for a countably generated sub-sigma-algebra, or a measurable partition represented by such a map. Sigma-algebras are considered modulo null sets where necessary.

For an invariant probability measure, disintegrate over the invariant sigma-algebra $\mathcal{I}$. On a standard probability space it can be represented, modulo null sets, by a countably generated measurable factor $\pi$. One justification for this countability is the separability of $L^2$: a countable dense family of $\mathcal{I}$-measurable functions generates $\mathcal{I}$ modulo null sets. The resulting conditional measures are the ergodic components:Their invariance is one of the permitted facts. It remains to prove that almost every component is ergodic, rather than simply invariant.

Choose a countable generating algebra $\mathcal{A}$ generating the Borel sigma-algebra of $X$. Apply the Birkhoff ergodic theorem simultaneously to its indicators. For $\mu$-almost every $z$ and all $A\in \mathcal{A}$,Disintegration transfers this common full-measure set to ${\mu }_y$-almost every $z$, for $\nu$-almost every $y$. On that fiber the conditional expectation is constant and equals ${\mu }_y(A)$. Thus, simultaneously for every $A\in \mathcal{A}$,Fix such a $y$, for which ${\mu }_y$ is also invariant. Applying the Birkhoff ergodic theorem to the system with measure ${\mu }_y$ identifies these limits withFinite linear combinations of the algebra's indicators are dense in $L^2({\mu }_y)$, by the Monotone class theorem. Since conditional expectation is an $L^2$ contraction, its projection onto invariant functions consequently sends every $L^2({\mu }_y)$ function to its constant integral. In particular, for any ${\mu }_y$-invariant set $D$,and so ${\mu }_y(D)\in \{0,1\}$. This proves almost every component is ergodic. The countable-test proof of ergodicity of conditional components avoids taking an invalid intersection of uncountably many full-measure sets.

For rotation by $1/5$, take the quotient map $\pi (x)=5x(\mathrm{mod}1)$. Its fibers are exactly the five-point orbits. The ergodic components of a rational circle rotation are thereforeEquivalently the component through $x$ is $\frac{1}{5}{\sum }_{j=0}^4{\delta }_{x+j/5}$. To check the disintegration, for bounded measurable $f$ use the change of variables on the five consecutive fifth-intervals:The rotation cyclically permutes the five atoms, preserving their equal masses. A set invariant modulo ${\mu }_y$ must contain all or none of this finite orbit, since every atom has positive mass. Thus each ${\mu }_y$ is ergodic. Finally, the invariant sigma-algebra is the pullback under $\pi$: a function invariant under translation by $1/5$ is constant on each orbit and is a measurable function of $5x(\mathrm{mod}1)$. This confirms that these are the conditional measures over $\mathcal{I}$, not merely some decomposition into invariant measures.

For an invertible probability measure-preserving system, an almost periodic observable is an $f\in L^2(\mu )$ whose Koopman operator orbit is totally bounded:Equivalently, for every $\epsilon >0$ it has a finite $\epsilon$-net in $L^2$. For a noninvertible system use the forward orbit $n\ge 0$.

A factor map $\phi :X\to Y$ satisfies ${\phi }_{\ast }\mu =\nu$ and $\phi T=S\phi$ almost everywhere. Disintegrate $\mu ={\int }_Y{\mu }_{y}d\nu (y)$ over this map and writefor the conditional L2 norm. A relatively almost periodic observable is an $f\in L^2(\mu )$ such that, for every $\epsilon >0$, there are finitely many $g_1,\dots ,g_r\in L^2(\mu )$ withThe center may depend on both $n$ and $y$; the finite list itself does not. Since $\mathbb{Z}$ is countable, the exceptional null sets can be combined. A compact extension of a measure-preserving system is a factor map for which these relatively almost periodic observables are dense in $L^2(\mu )$. This density definition should not be replaced by a claim that every $L^2$ observable already satisfies the uniform fiberwise finite-net condition.

For a proper factor map example, let $Y=\{-1,1{\}}^{\mathbb{Z}}$ carry independent fair coordinates and the two-sided Bernoulli shift $S$, with $(Sy{)}_j=y_{j+1}$. Setusing the product of the Bernoulli measure and the equal two-point measure. These are invertible measure-preserving systems, and the projection preserves measure and intertwines the transformations, so it is a two-to-one factor map.

Take $f(y,i)=y_0$. Then $U_T^{n}f(y,i)=y_n$. Distinct coordinates are independent and have mean zero and variance one, henceIts orbit has no finite sufficiently small net. Thus the Bernoulli coordinate observable is not almost periodic in global $L^2$. In contrast, the conditional measure over $y$ is the equal measure on $(y,0),(y,1)$, and $U_T^{n}f$ is constant on that fiber with value $y_n\in \{-1,1\}$. The two centers $g_1=1,g_2=-1$ give zero error on every fiber for every $n$. Therefore $f$ is a relatively almost periodic observable for this map.

The example is also a finite-fiber compact extension. For a bounded observable on $X$, each orbit value on a fiber is a pair in a fixed bounded subset of ${\mathbb{C}}^2$, which has a finite net. Its centers can be chosen as functions constant in $y$, with prescribed values on the two labels. Bounded observables are dense in $L^2$, so the extension is compact. The Bernoulli shift is mixing: cylinder events involving disjoint coordinate sets are independent for all sufficiently large shifts, and approximation by cylinder events extends this to arbitrary events. Thus $S$ and $S^2$ are ergodic. If an invariant function on $X$ is written $a(y)+(-1{)}^{i}b(y)$, invariance gives $a\circ S=a$ and $b\circ S=-b$. The first function is constant; the second is $S^2$-invariant and hence constant, and then its sign equation forces zero. This proves that the example's source is also ergodic.

For the positive-subset assertion, interpret the printed $X^{\prime }$ as $X$, or as an invariant full-measure domain of the factor map; the prime is otherwise undefined and does not change the measure-theoretic argument.

First establish conditional norm covariance under a factor map. Invariance of the measures and uniqueness in the disintegration theorem for a probability measure giveon common full-measure sets. For the first equality, $y\mapsto T_{\ast }{\mu }_{S^{-1}y}$ is another disintegration over the same fibers: it is supported on ${\phi }^{-1}\{y\}$ and integrates to $\mu$. Uniqueness identifies it with ${\mu }_y$, and iteration gives the norm formula.

Two elementary facts explain localization of relative almost periodicity to base sets. If $f$ is a relatively almost periodic observable and $C\subseteq Y$ is measurable, then $f{1}_C\circ \phi$ is also relatively almost periodic. Indeed, on a fiber,When that indicator is one, use a center for $U_T^{n}f$; when it is zero, use the additional center $0$. Secondly, relative almost periodicity is closed under convergence in the uniform conditional L2 normIf this norm of $f-f_j$ tends to zero, covariance bounds the corresponding error between every pair of orbit values by the same number. A finite $\epsilon /2$-net for one sufficiently close $f_j$ is consequently an $\epsilon$-net for $f$.

Now take $f=1_A$. The compact extension supplies relatively almost periodic observables $f_j$ withPut $e_j(y)=\Vert f_j-f{\Vert }_y$. The Tonelli theorem givesHence $e_j(y)\to 0$ for $\nu$-almost every $y$. Since ${\int }_Y{\mu }_y(A)d\nu =\mu (A)>0$, choose $\eta >0$ for whichhas positive measure. The Egorov theorem gives a measurable $C\subseteq E$ of positive measure on which $e_j\to 0$ uniformly. DefineThenFor $h_j=f_j{1}_C\circ \phi$, the localization fact makes every $h_j$ relatively almost periodic, whileThe conditional-norm closure fact therefore proves $1_B$ is a relatively almost periodic observable, with $B\subseteq A$ and positive measure. This is the positive-measure almost periodic indicator in a compact extension. No ergodic component or uncountable intersection of exceptional sets is needed in this construction; it in fact works under the same disintegration and compactness hypotheses without using the assumed ergodicity.

Use the same formula for a measurable representative of $u_0$. Dilation preserves null sets and gives $\Vert u{\Vert }_{L^{\infty }}=\Vert u_0{\Vert }_{L^{\infty }}$. To check the weak formulation, change variables $x=e^{t}y$ and set $\psi (t,y)=e^t\phi (t,e^{t}y)$. ThenThe spacetime integral becomes $\int u_0(y){\psi }_t(t,y)dydt=-\int u_0(y)\phi (0,y)dy$, as required.

For uniqueness, take any bounded weak solution and write $v(t,y)=u(t,e^{t}y)$. Choosing $\phi (t,x)=e^{-t}\psi (t,e^{-t}x)$ in the weak formulation givesFor tensor test functions $\psi (t,y)=\eta (t)\theta (y)$, this says that $\int v(t,y)\theta (y)dy$ is distributionally constant and has value $\int u_0\theta$. A countable dense family of spatial test functions identifies $v=u_0$ almost everywhere. Thus the displayed solution is the unique bounded weak solution.

For $I=(-a,a)$, the first-order Sobolev space isThe weak derivative $v=u^{\prime }$ is unique as an equivalence class modulo equality almost everywhere. We use the standard Sobolev norm $\Vert u{\Vert }_{H^1(I)}^2=\Vert u{\Vert }_{L^2(I)}^2+\Vert u^{\prime }{\Vert }_{L^2(I)}^2$. Pointwise values of an arbitrary representative have no intrinsic meaning; the one-dimensional Sobolev representative supplies the continuous representative used later.

Reflect $u$ across the two endpoints of $I$ into a slightly larger interval, giving an $H^1$ extension, and multiply by a smooth cutoff function equal to one near $\overline{I}$. Convolve the resulting function on $\mathbb{R}$ with a mollifier. The convolutions $u_{\epsilon }$ are smooth; differentiation commutes with convolution of the weak derivative. Continuity of translations in $L^2$ proves both $u_{\epsilon }\to u$ and $u_{\epsilon }^{\prime }\to u^{\prime }$ in $L^2(I)$, hence convergence in the Sobolev norm.

To obtain convergence almost everywhere, select a sequence with $\Vert u_n-u{\Vert }_2\le 2^{-n}$. Then $\int {\sum }_n|u_n-u{|}^2<\infty$, so the summands tend to zero almost everywhere. Thus a smooth approximating sequence converges both in $H^1$ and almost everywhere. This is density of smooth functions in a Sobolev space; smooth functions need not vanish at the endpoints.

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.