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For real random variables $X,Z$, if $Z$ is measurable with respect to the sigma-algebra generated by a random variable $X$, there is a Borel measurable function $g:\mathbb{R}\to \mathbb{R}$ with $Z=g(X)$. To prove the bounded case, approximate $Z$ pointwise by finite-valued simple functions $Z_m$. Each level set is $X^{-1}(B)$ for a Borel set $B$, so each approximation factors as $Z_m=g_m(X)$ with $g_m$ a Borel measurable function. Apply scalar clipping to $g_m$ with a common bounded interval and use its limit superior to obtain $g$. For unbounded finite $Z$, apply the bounded case to the one-to-one bounded continuous function $t\mapsto \arctan t$, then invert on $(-\pi /2,\pi /2)$ and assign arbitrary values elsewhere. Exact measurability gives exact factorization; assumptions after adjoining null sets give an almost surely equal version.

Varadhan's lemma states that if $U_n$ has a large deviation principle with good rate function $K$ and large-deviation speed $a_n$, then for every bounded continuous real function $h$,For a continuous function unbounded above, the same conclusion holds under the exponential tail extension of Varadhan's lemma, for example when the weighted contribution from $h>M$ is superexponentially negligible as $M\to \infty$.

The scalar clipping map $g(m)=\min (b,\max (a,m))$ is continuous. The rate function under scalar clipping has endpoint costs ${\inf }_{m\le a}J(m)$ and ${\inf }_{m\ge b}J(m)$. Since $a<\mu <b$ and the costs decrease towards $\mu$ from the left and increase to its right, both infima are attained at the corresponding endpoints. Thus the good rate function of $Z_n$ equals $J(z)$ on $[a,b]$ and is infinite outside. The function $\log z$ is bounded and continuous on that compact interval, soOn $[a,\mu ]$, the derivative of $\log z-I(z)$ is $2/z-\lambda >0$, so the maximum there is $\log \mu$. On $[\mu ,b]$, the derivative is $(k+1)/z-k\lambda$, with negative second derivative. Its unconstrained maximizer isIt lies in $[\mu ,b]$ whenever $k\ge (\lambda b-1{)}^{-1}$. The right-hand objective increases initially at $\mu$, so this maximizer dominates the whole left-hand interval. Substitution gives the limit for high moments of a clipped minimum of exponential sample meansThe moment here is the expectation of the $n$th power. If the printed notation were instead read as $(\mathbb{E}{Z}_n{)}^n$, its logarithmic limit would be $\log \mu =-\log \lambda$, since boundedness and convergence in probability give $\mathbb{E}{Z}_n\to \mu$.

For $m\ge 1$, fix signs $x_1,\dots ,x_m$. The constraints $X_j=x_j$ leave exactly two choices for the $(m+1)$ signs $Y_0,\dots ,Y_m$: choose $Y_0$ freely, then set $Y_1=Y_0{x}_1$ and $Y_j=x_{j-1}{x}_j$ for $2\le j\le m$. Each input tuple has probability $2^{-(m+1)}$ by independence. ConsequentlyEvery finite subfamily is a marginal distribution of one such initial segment, so the entire sequence is independent, with fair sign probability distributions.

Using the usual everywhere sign-valued representatives, for every $n\ge 1$ we haveThe right side is measurable with respect to $\sigma (\mathcal{Y},{\mathcal{T}}_n)$. Therefore $Y_0$ is measurable with respect to their intersection $\mathcal{G}$.

Write $\mathcal{T}={\bigcap }_n{\mathcal{T}}_n$. The Kolmogorov zero-one law applies to the independent random variables $(X_n)$, so every event $C\in \mathcal{T}$ has probability zero or one. Also $Y_0$ is independent of $\mathcal{Y}$: this follows first for finite intersections of sign events from the given independence, then for their generated sigma-algebra by part (a).

The sets $D\cap C$ with $D\in \mathcal{Y}$ and $C\in \mathcal{T}$ form a pi-system generating $\mathcal{H}$. If $\mathbb{P}(C)=0$, the event $D\cap C$ has probability zero and is independent of every event in $\sigma (Y_0)$. If $\mathbb{P}(C)=1$, it differs from $D$ by a null set, so the same independence follows from that of $Y_0$ and $\mathcal{Y}$. Another application of part (a) provesIn particular $\mathcal{H}$ cannot contain the nontrivial event $\{Y_0=1\}$, whereas $\mathcal{G}$ does. Thus adjoining a sigma-algebra need not commute with a decreasing intersection.

There is a small version qualification if the assumptions specify only the probability distributions, allowing exceptional values other than signs. Put $q_n=X_n/(Y_1\cdots Y_n)$ when the denominator is nonzero, and $q_n=0$ otherwise. For $k\ge n$, $q_k$ is measurable with respect to $\sigma (\mathcal{Y},{\mathcal{T}}_n)$. The limit superior of $(q_k)$, after scalar clipping to $[-1,1]$, is therefore $\mathcal{G}$-measurable and equals $Y_0$ almost surely. Literal measurability of an arbitrary version of $Y_0$ need not follow without choosing sign-valued representatives or adjoining the ambient null sets. All the independence conclusions are unaffected.

We prove the Doob-Dynkin lemma here by simple function approximation. Choose $C<\infty$ such that $|Z|\le C$ everywhere, and letEach $Z_m$ is a finite-valued $\sigma (X)$-measurable function and $Z_m\to Z$ pointwise. Write its level sets as $A_{m,j}$, with associated values $z_{m,j}$. By part (i), choose Borel sets $D_{m,j}$ with $A_{m,j}=X^{-1}(D_{m,j})$. Define the Borel measurable functionsAlthough the chosen $D_{m,j}$ need not partition all of $\mathbb{R}$, their inverse images do partition $\Omega$, so $g_m(X)=Z_m$ exactly.

Apply scalar clipping to each $g_m$ with bounds $[-C-1,C+1]$, preserving this equality, and define $g(x)={\mathrm{limsup}}_m{g}_m(x)$. A bounded limit superior of Borel measurable functions is a finite Borel measurable function. At every $\omega \in \Omega$ the limit of a sequence exists and givesThus the factorization is exact for a bounded, exactly $\sigma (X)$-measurable function. If hypotheses are instead given only up to null sets, the corresponding conclusion is a factorization holding almost surely after choosing a representative.

Choose $Z_{\infty }$ as the pointwise limit of a sequence of $(Z_n)$ where it exists, and as zero otherwise. The convergence event and every $Z_n$ are $\sigma (X)$-measurable, so this is an exactly $\sigma (X)$-measurable version, bounded by $L$. Part (a)(ii), the Doob-Dynkin lemma, supplies a Borel measurable function $g$ on $[0,1]$ such that $Z_{\infty }=g(X)$. Applying scalar clipping to $g$ with bounds $[-L,L]$ preserves that equality and ensures $|g(u)|\le L$ everywhere.

Part (c)(ii) and the conditional expectation formula from part (b)(iv) now giveSince the indices $n$ form a countable family, these identities may be taken on one common full-probability event.

For fixed $n,k$, both the slope and the displayed average are constant on the atom $A_{n,k}$, whose probability is $2^{-n}>0$. An equality holding almost surely of these constants forces the deterministic equalitySum over $0\le k<j$. The telescoping sum givesThus the desired identity holds at every dyadic endpoint, including $1$, without evaluating the undefined slope at $X=1$.

Finally, given $x\in [0,1]$, take dyadic endpoints $x_n\to x$. The Lipschitz condition implies $|f(x_n)-f(x)|\le L|x_n-x|$, and the bound on the Borel measurable function $g$ givesPassing to the limit of a sequence provesThis is a martingale proof of the integral representation of a Lipschitz function. It constructs a bounded Lebesgue integrable function representing the increments of $f$, and in particular proves absolute continuity of $f$. No prior assumption that $f$ has a derivative is needed.

We first establish the lower truncation identity as well. The pair $(-X,-Y)$ has the same hypotheses as $(X,Y)$, because conditional expectation is linear and negation preserves equality of probability distributions. Apply part (ii) to this pair and multiply by $-1$ to getFor every real $t$,Using this identity, part (ii), the lower truncation identity, and linearity of conditional expectation givesIn particular the argument does not assume that a function obtained by scalar clipping is a convex function or a concave function.

Write $C_a(t)=(t\wedge a)\vee (-a)$. The bounded random variables $C_a(X)$ and $C_a(Y)$ have the same probability distribution, and the displayed identity identifies the latter as the conditional expectation of the former. They are square-integrable, so part (i) gives $C_a(X)=C_a(Y)$ almost surely. Take $a=1,2,\dots$ and discard the countable union of exceptional null sets. At every remaining outcome, let $a\to \infty$ to obtainThus conditional expectation preserving a distribution forces equality even when only integrability is assumed.

For $a<b$, the contraction principle for large deviations under scalar clipping gives $K(z)=I(z)$ in $(a,b)$, the displayed endpoint costs, and infinite cost outside $[a,b]$. The endpoint costs can differ from $I(a)$ and $I(b)$ because clipping collects a whole input tail into a single output atom.