# Convergence in probability to a non-measurable limit

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.

## 2 Answers

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.

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.

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
• I think you are missing from the hypotheses of theorem 1 that $X_n \leq Z_n$ on $E_n$, no? With that change, however, theorem 1 looks good. I'll have to look over the rest a bit later, but nice work! – user139388 Apr 3 '14 at 20:18
• @user139388: I don't think so. $X_n = \mu_n$, $Y = \mu$, $Z_n = \overline{\mu}_n$. – Evan Aad Apr 3 '14 at 20:22
• Yes, I see. Okay, again, great work! Sorry I wasn't more helpful – user139388 Apr 3 '14 at 20:24
• @user139388: Thanks. You were very helpful. The mere fact that you took interest in my question and made an honest effort to solve it encouraged me to continue to think hard about the problem until I eventually came up with this solution. So thank you! – Evan Aad Apr 3 '14 at 20:27

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.

• I have updated my question to address your questions concerning notation. – Evan Aad Apr 1 '14 at 7:16
• Why does the sequence $a_n := \sup_{E_n}\{\overline{\mu}_n(\omega) - \mu(\omega)\}$ converge to $0$? – Evan Aad Apr 1 '14 at 7:26
• Okay, thanks Evan. Actually, I think its not necessarily true (about $a_n$). Let me think again on it ... – user139388 Apr 1 '14 at 7:30
• By the way, I think you might be able to use the Skorohod representation theorem for the second part. – user139388 Apr 1 '14 at 7:42
• I tried another idea, but still feel shaky. Please check it! – user139388 Apr 1 '14 at 8:03