Convergence in probability to a non-measurable limit

Question

Let $(\Omega, \mathcal{F}, P)$ be a probability space. Denote the Borel field on $\mathbb{R}$ by $\mathcal{B}$. Let $\mu: \Omega \rightarrow [0,\infty)$ be a not-necessarily-measurable function and, for every $n \in \mathbb{N}_1 := \{1, 2, \dots\}$, let $\mu_n:\Omega \rightarrow \mathbb{R}$ be $\mathcal{F}/\mathcal{B}$-measurable.

1. For every $n \in \mathbb{N}_1$ let $\overline{\mu}_n:\Omega \rightarrow [0,\infty)$ be a not-necessarily-measurable function, such that $\mu, \mu_n \leq \overline{\mu}_n$ and $\lim_{n \rightarrow \infty}\overline{\mu}_n = \mu$, point-wise, and let $E_n \in \mathcal{F}$ be such that $\lim_{n \rightarrow \infty}P(E_n) = 1$ and such that for every $\omega \in E_n$, $\mu(\omega) \leq \mu_n(\omega)$.

   Does $\mu_n$ converge in probability to $\mu$, in the sense that for every $\varepsilon \in (0,\infty)$ there is a sequence of events $(F_1, F_2, \dots) \in \mathcal{F}^\infty$, such that $\lim_{n \rightarrow \infty} P(F_n) = 1$ and such that for all $n \in \mathbb{N}_1$ and all $\omega \in F_n$, $|\mu_n(\omega) - \mu(\omega)| < \varepsilon$?

2. Suppose that for every $\varepsilon \in (0,\infty)$ there is a sequence of events $(F_1, F_2, \dots) \in \mathcal{F}^\infty$, such that $\lim_{n \rightarrow \infty}P(F_n) = 1$ and such that for all $n \in \mathbb{N}_1$ and all $\omega \in F_n$, $|\mu_n(\omega) - \mu(\omega)| < \varepsilon$. Is $\mu$ necessarily $\mathcal{F}/\mathcal{B}$-measurable?

Remark: The motivation for this question is that these two claims (if I understand correctly, as I have paraphrased a little) are stated without proof in the paper Le mouvement Brownien plan (1940) by Paul Lévy (p. 533) as part of Lévy's proof of the fact that a planar Brownian motion's area is $0$ almost surely. $\mu$ is the area of the Brownian motion, $\mu_n$ are measurable approximations to the area and $\overline{\mu}_n$ are non-measurable, convergent approximations. The point is to demonstrate that $\mu$ is measurable as the limit in probability of a sequence of measurable functions.

Answer 1

Theorem 1 below proves that under the conditions of OP's question #1, $\mu$ is measurable. This result is what OP was really after, and therefore in light of theorem 1, answering OP's two questions becomes unnecessary for OP's purposes. Nevertheless, I have included an (affirmative) answer to question #2 (theorem 6), since it holds independent interest, in my opinion. Definitions 2 & 3 and lemmas 4 & 5 are used in theorem 6's proof. The references [A] and [K] are listed in the section "Works cited" at the bottom of this post.

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Theorem 1 (If a sequence of random variables is "squeezed" under a convergent sequence of arbitrary functions, the limit is measurable) Let $X_1, X_2, \dots$ be a sequence of random variables over the probability space $(\Omega, \mathcal{F}, P)$, and let $Y, Z_1, Z_2, \dots : \Omega \rightarrow \mathbb{R}$ be arbitrary (i.e. not necessarily measurable). Suppose that

1. For all $n \in \mathbb{N}_1$, $X_n, Y \leq Z_n$,
2. $\lim_{n \rightarrow \infty} Z_n = Y$,
3. There is a sequence of events $E_1, E_2, \dots \in \mathcal{F}$ with $\lim_{n \rightarrow \infty} P(E_n) = 1$, such that for every $n \in \mathbb{N}_1$ and every $\omega \in \Omega$, $X_n(\omega) \geq Y(\omega)$.

Then there is a random variable $X$ and an event $E \in \mathcal{F}$ with $P(E) = 1$, such that $$ E \subseteq \{X = Y\} $$

Proof

Define $X := \limsup_{n \rightarrow \infty} X_n$. $X$, which takes values in $\mathbb{R} \cup \{\pm\infty\}$, is a random variable ([K] Theorem 1.92, p. 40). Define $E := \limsup_{n \rightarrow \infty} E_n$. $E \in \mathcal{F}$. By Fatou's lemma, $P(E) \geq \limsup_{n \rightarrow \infty} P(E_n) = 1$. Let $\omega \in E$. Then on the one hand, $$ X(\omega) \leq \limsup_{n \rightarrow \infty} Z_n(\omega) = Y(\omega) $$ But on the other hand, there is a strictly ascending sequence $(n_1, n_2, \dots) \in \mathbb{N}_1^\infty$, such that $\omega \in \bigcap_{k = 1}^\infty E_{n_k}$. For every $k \in \mathbb{N}_1$, $X_{n_k}(\omega) \geq Y(\omega)$, so $X(\omega) \geq Y(\omega)$. In conclusion, $X(\omega) = Y(\omega)$.

Q.E.D.

Definition 2 (Convergence in probability to an arbitrary function) Let $\mathbf{X} = (X_1, X_2, \dots)$ be a sequence of random variables over the probability space $(\Omega, \mathcal{F}, P)$ and let $Y:\Omega \rightarrow \mathbb{R}$ be some, not necessarily measurable, function. We say that $\mathbf{X}$ converges in probability to $Y$, or, in symbols, $$ X_n \overset{P}{\longrightarrow} Y $$ iff for every $\varepsilon \in (0,\infty)$ there is some sequence $(E_1, E_2, \dots) \in \mathcal{F}^\infty$ with $\lim_{n \rightarrow \infty} P(E_n) = 1$, such that for all $n \in \mathbb{N}_1$ and for all $\omega \in E_n$, $$ |X_n(\omega) - Y(\omega)| < \varepsilon $$

Definition 3 (Cauchy sequence in probability) Let $\mathbf{X} = (X_1, X_2, \dots)$ be a sequence of random variables over the probability space $(\Omega, \mathcal{F}, P)$. $\mathbf{X}$ is Cauchy in probability iff for every $\varepsilon \in (0,\infty)$ $$ \lim_{N \rightarrow \infty}\sup_{m, n \in \{N, N + 1, \dots\}}P(|X_m - X_n| < \varepsilon) = 1 $$

Lemma 4 (A sequence convergent in probability to an arbitrary function is Cauchy in probability) Let $\mathbf{X} = (X_1, X_2, \dots)$ be a sequence of random variables over the probability space $(\Omega, \mathcal{F}, P)$ and let $Y:\Omega \rightarrow \mathbb{R}$ be some, not necessarily measurable, function. If $\mathbf{X}$ converges in probability to $Y$, $\mathbf{X}$ is Cauchy in probability.

Proof

Let $\delta, \varepsilon \in (0,\infty)$ and let $E_1, E_2, \dots \in \mathcal{F}$ be such that $\lim_{i \rightarrow \infty} P(E_i) = 1$ and such that for all $i \in \mathbb{N}_1$ and for all $\omega \in E_i$, $|X_i(\omega) - Y(\omega)| < \varepsilon / 2$. Choose some $N \in \mathbb{N}_1$, such that for all $i \in \{N, N + 1, \dots\}$, $P(E_i) > 1 - \delta / 2$ and let $m, n \in \{N, N+1, \dots\}$. Define $E := E_m \cap E_n$. We have $P(E) \geq P(E_m) + P(E_n) - 1 > 1 - \delta$ and for every $\omega \in E$, $$ |X_m(\omega) - X_n(\omega)| \leq |X_m(\omega) - Y(\omega)| + |Y(\omega) - X_n(\omega)| < \varepsilon $$ so that $$ \sup_{m, n \in \{N, N+1, \dots\}} P(|X_m - X_n| > \varepsilon) \geq 1 - \delta $$ Q.E.D.

Lemma 5 (A sequence of functions that is Cauchy in probability, converges in probability to a measurable function) Let $\mathbf{X} = (X_1, X_2, \dots)$ be a sequence of random variables over the probability space $(\Omega, \mathcal{F}, P)$ and suppose that $\mathbf{X}$ is Cauchy in probability. Then there is some random variable $Y : \Omega \rightarrow \mathbb{R}$, such that $$ X_n \overset{P}{\longrightarrow} Y $$

Proof

The first part of the proof of theorem 2.3.5 in [A] (p. 97-98) shows that there is some random variable $Y : \Omega \rightarrow \mathbb{R}$ and some strictly ascending sequence $(n_1, n_2, \dots) \in \mathbb{N}_1^\infty$, such that $$ X_{n_k} \overset{\textrm{a.s.}}{\longrightarrow} Y $$ hence $$ X_{n_k} \overset{P}{\longrightarrow} Y $$

Let $\delta, \varepsilon \in (0,\infty)$, and let $K, L \in \mathbb{N}_1$ be such that

1. For all $k \in \{K, K + 1, \dots\}$, $$ P(|X_{n _k} - Y| < \varepsilon/2) > 1 - \delta/2 $$
2. For all $m, n \in \{L, L + 1, \dots\}$, $$ P(|X_m - X_n| < \varepsilon/2) > 1 - \delta/2 $$

Define $N := \max(n_K.L)$. Then for all $n \in \{N, N + 1, \dots\}$, $$ \begin{align} P(|X_n - Y| < \varepsilon) & \geq P(|X_n - X_{n_K}|, |X_{n_K} - Y| < \varepsilon/2) \\ & \geq P(|X_n - X_{n_K}| < \varepsilon/2) + P(|X_n - X_{n_L}| < \varepsilon/2) - 1 \\ & > 1 - \delta \end{align} $$ Q.E.D.

Theorem 6 (A limit in probability is measurable)

Let $\mathbf{X} = (X_1, X_2, \dots)$ be a sequence of random variables over the probability space $(\Omega, \mathcal{F}, P)$ and let $Y:\Omega \rightarrow \mathbb{R}$ be some, not necessarily measurable, function, such that $\mathbf{X}$ converges in probability to $Y$. Then there exists some random variable $Z: \Omega \rightarrow \mathbb{R}$ and some $B \in \mathcal{F}$ with $P(B) = 0$, such that $$ \{Y \neq Z\} \subseteq B $$

Proof

By the lemmas above, there exists some random variable $Z:\Omega \rightarrow \mathbb{R}$ such that $$ X_n \overset{P}{\longrightarrow} Z $$

Define $A := \{Y \neq Z\}$ ($A$ is not necessarily $\in \mathcal{F}$.) I will show that there exists some $B \in \mathcal{F}$ with $P(B) = 0$, such that $A \subseteq B$.

To every $k \in \mathbb{N}_1$, let $C_k, D_k \in \mathcal{F}$ be such that $P(C_k), P(D_k) > 1 - 2^{- (k + 1)}$ and such that for all $\omega \in C_k$, $\omega' \in D_k$, $$ |Y(\omega) - X_k(\omega)|, |X_k(\omega') - Z(\omega')| < \frac{1}{2k} $$

Define $E_k := C_k \cap D_k$, $F_k := E_k^c$. Then $P(E_k) \geq P(C_k) + P(D_k) - 1 > 1 - 2^k$ (and therefore $P(F_k) \leq 2^{-k}$) and for all $\omega \in E_k$, $$ |Y(\omega) - Z(\omega)| < \frac{1}{k} $$

Define $B := \limsup_{k \rightarrow \infty} F_k$. Then $B \in \mathcal{F}$ and by the Borel-Cantelli lemma, $P(B) = 0$. Let $\omega \in A$ and set $\delta_\omega := |Y(\omega) - Z(\omega)|$. Then $\delta_\omega > 0$. Define $K := \lceil\delta_\omega^{-1}\rceil$. Then for all $k \in \{K, K + 1, \dots\}$, $\omega \in F_k$, and therefore $\omega \in \liminf_{k \rightarrow \infty} F_k$. Hence $$ A \subseteq \liminf_{k \rightarrow \infty} F_k \subseteq B $$

Q.E.D.

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WORKS CITED

• [A] Ash, Robert B. and Doléans-Dade, Catherine A. Probability & Measure Theory. 2nd edition. Academic Press, 2000
• [K] Klenke, Achim. Probability Theory, A Comprehensice Course. Springer, 2008

Answer 2

Let $A_n(\epsilon) = \{|\mu_n - \mu| \geq \epsilon \}$. If $\mu$ were measurable $$ P[\mu_n \not\to \mu ] \leq P[\exists N, \forall n\geq N, |\mu_n - \mu| \geq \epsilon ] = P[\liminf A_n(\epsilon)] \leq \liminf P[A_n(\epsilon)], $$ (if not, it is hard to talk about converging in probability anyway since the sets $\{ |\mu_n - \mu| \geq \epsilon \}$ won't be measurable).

Now $P[A_n(\epsilon)] = P[A_n(\epsilon)\cap E_n] + P[A_n(\epsilon)\cap E_n^c]$ and taking the $\liminf$ shows that $$ \liminf P[A_n(\epsilon)] \leq \liminf P[A_n(\epsilon)\cap E_n] + \liminf P[E_n^c] = \liminf P[A_n(\epsilon)\cap E_n]. $$ On the set $A_n(\epsilon)\cap E_n$, $\bar{\mu}_n - \mu \geq \epsilon$ is true, but this must happen only finitely many times since $\bar{\mu}_n \to \mu$ at all $\omega$. so $$ \liminf P[A_n(\epsilon)\cap E_n] \leq P[\limsup A_n(\epsilon)\cap E_n] = 0. $$ So I think almost sure convergence holds.