Generalizing Scheffe's Lemma using only Convergence in Probability

Question

I thought about this question recently, because we accidentally stated this lemma with convergence in probability instead of the usual almost sure convergence as an exercise. The usual proof with Fatou's lemma does not work in this case.

It turns out that you can in fact generalize Scheffe's lemma to the following

Generalized Scheffe's Lemma

Assume that $(X_n)_{n\in\mathbb{N}}\subset L^1 $ converges in probability to $X_\infty\in L^1$. Then the following statements are equivalent:

1. $\mathbb{E}[|X_{n}|]\to \mathbb{E}[|X_{\infty}|]<\infty$, as $n\to \infty$.
2. For all $\epsilon>0$ we have $\limsup_{n\to\infty} \mathbb{E}[|X_n|\mathbb{1}_{|X_\infty-X_n|> \epsilon}] \le \epsilon$
3. $\{X_\infty, X_1, X_2,\dots\}$ are uniformly integrable
4. $X_n\to X_{\infty}$ in $L^1$, as $n\to \infty$

I am not sure if this result is new (probably not) but I wanted it to become easier to find. So here is a proof.

Answer 1

Proof

(i) $\Rightarrow$ (ii): Fix $\epsilon>0$ and choose any $\eta>0$. Then there exists $n_0\in\mathbb{N}$ such that

\begin{align}\label{generalized scheffe (i)->(ii)1} \mathbb{E}[|X_n|] - \mathbb{E}[|X_\infty|] \le \tfrac{\eta}2 \quad \forall n\ge n_0. \end{align}

Since $\{X_\infty\}$ is uniformly integrable, there exists $\delta(\eta)>0$ such that \begin{align*} \mathbb{E}[|X_\infty| 1_{A}] \le \tfrac\eta2 \quad \forall A\in\mathcal{A} : \mathbb{P}(A) < \delta(\eta). \end{align*} Since $X_n\to X_\infty$ in probability, there therefore exists $n_1\in\mathbb{N}$ such that for all $n\ge n_1$ \begin{align}\label{generalized scheffe (i)->(ii)2} \mathbb{P}(|X_\infty - X_n| > \epsilon) \le \delta(\eta) \implies \mathbb{E}[|X_\infty| 1_{|X_\infty - X_n| > \epsilon}] \le \tfrac\eta2 \end{align} Now we plug things together. We have for $n\ge\max\{n_0, n_1\}$ \begin{align*} \mathbb{E}[|X_n| 1_{|X_\infty-X_n|>\epsilon}] &= \mathbb{E}[|X_n|] - \mathbb{E}[|X_n| 1_{|X_\infty-X_n|\le\epsilon}]\\ &= \underbrace{\mathbb{E}[|X_n|] - \mathbb{E}[|X_\infty|]}_{ \le \tfrac\eta2 } + \underbrace{ \mathbb{E}[(|X_\infty| - |X_n|)1_{|X_\infty-X_n|\le\epsilon}] }_{ \begin{aligned} &\le \mathbb{E}[|X_\infty - X_n|1_{|X_\infty-X_n|\le\epsilon}]\\ &\le \epsilon \end{aligned} } + \underbrace{\mathbb{E}[|X_\infty|1_{|X_\infty-X_n|>\epsilon}]}_{ \le \tfrac\eta2 } \end{align*} This implies the claim. The first sum exploits the fact that the mass of the random variables has to be similar. So we can not concentrate mass on a small event such as $X_n$ being far from $X_\infty$. Which is what examples of convergence in probability without $L^1$ convergence exploit.

(ii) $\Rightarrow$ (iii): We want to show \begin{align*} \lim_{M\to\infty} \sup_n \mathbb{E}[|X_n|1_{|X_n|> M}] = 0. \end{align*} So fix some $\delta>0$. We select $\epsilon := \tfrac\delta4$ and then select $n_0\in\mathbb{N}$ such that for all $n\ge n_0$ we have \begin{align}\label{convergence ui} \mathbb{E}[|X_n|1_{|X_\infty - X_n| > \epsilon}] \le \epsilon + \tfrac\delta4. \end{align} As $\{X_1, \dots, X_{n_0}\}$ is a finite set it is uniformly integrable, and there exists $M_0$ such that \begin{align}\label{finite case} \sup_{0\le n\le n_0} \mathbb{E}[|X_n| 1_{|X_n| > M}] \le \delta \quad \forall M\ge M_0. \end{align} Similarly there exists $M_1$, such that \begin{align}\label{X_infty is ui} \mathbb{E}[|X_\infty| 1_{|X_\infty|>M-\epsilon}] \le \tfrac\delta4 \quad \forall M\ge M_1 \end{align} To put things together, note that we always have \begin{align}\label{indicator functions} 1_{|X_n| > M} \le 1_{|X_\infty - X_n| > \epsilon} + 1_{|X_\infty|>M-\epsilon}1_{|X_\infty-X_n|\le \epsilon}. \end{align} This implies for all $M\ge \max\{M_0, M_1\}$ \begin{align*} &\sup_{n\in\mathbb{N}} \mathbb{E}[|X_n| 1_{|X_n| > M}]\\ &\le \max\Big\{ \underbrace{\sup_{0\le n\le n_0} \mathbb{E}[|X_n| 1_{|X_n| > M}]}_{\le \delta}, \sup_{n\ge n_0} \underbrace{\mathbb{E}[|X_n| 1_{|X_\infty - X_n|>\epsilon}]}_{ \le \tfrac\delta2 } + \underbrace{\mathbb{E}[|X_n| 1_{|X_\infty|>M-\epsilon}1_{|X_\infty - X_n|\le\epsilon}]}_{ \begin{aligned} &\le \mathbb{E}[|X_\infty| 1_{|X_\infty|>M-\epsilon}] + \epsilon\\ &\le \tfrac\delta2 \end{aligned} } \Big\} \end{align*}

(iii) $\Rightarrow$ (iv): Fix some $\epsilon >0$, then due to uniform integrability there exists some $\delta >0$, such that $\mathbb{P}(A)<\delta$ implies for all $n$ \begin{align}\label{ui result} \mathbb{E}[|X_n| 1_{A}] \le \epsilon \quad \text{and}\quad \mathbb{E}[|X_\infty|1_{A}] \le \epsilon. \end{align} Now we choose $n_0\in \mathbb{N}$ such that \begin{align*} \mathbb{P}(|X_\infty - X_n| > \epsilon) \le \delta \quad \forall n\ge n_0. \end{align*}
With the previous result, this implies for all $n\ge n_0$ \begin{align*} \mathbb{E}[|X_\infty - X_n|] \le \underbrace{\mathbb{E}[|X_\infty - X_n|1_{|X_\infty - X_n|\le \epsilon}]}_{\le \epsilon} + \underbrace{ \mathbb{E}[|X_\infty| 1_{|X_\infty - X_n| > \epsilon}] }_{\le \epsilon} + \underbrace{ \mathbb{E}[|X_n| 1_{|X_\infty - X_n| > \epsilon}] }_{\le \epsilon} \end{align*}

(iv) $\Rightarrow$ (i): This finally follows from the reverse triangle inequality \begin{align}\label{generalized scheffe: reverse triangle inequality} \big\lvert|x|-|y|\big\rvert\le |x-y|, \end{align} and Jensen's inequality \begin{align*} \big\lvert \mathbb{E}\big[\lvert X_n \rvert\big] - \mathbb{E}\big[\lvert X_{\infty}\rvert\big] \big\rvert &= \big\lvert \mathbb{E}\big[\lvert X_n \rvert- \lvert X_{\infty}\rvert\big] \big\rvert\\ &\le \mathbb{E}\big[ \big\lvert \lvert X_n\rvert - \lvert X_{\infty}\rvert \big\rvert\big]\\ &\le \mathbb{E}\big[\lvert X_n - X_{\infty}\rvert\big] \to 0, \; (n \to \infty). \end{align*}

Answer 2

I'll demonstrate how to show 1 implies 4 only using the a.s. version of this theorem. Suppose $E(|X_n|) \to E(|X_\infty|)$. Let $(X_{n_k})$ be an arbitrary subsequence of $(X_n)$. Since $X_{n_k} \to X_{\infty}$ in probability, there is a subsequence $(X_{{n_k}_j})$ such that $X_{{n_k}_j} \to X_{\infty}$ a.s. Since $E(|X_{{n_k}_j}|) \to E(|X_{\infty}|)$, the a.s. version of the theorem implies that $X_{{n_k}_j} \to X_{\infty}$ in $L^1$. Since $X_{n_k}$ was an arbitrary subsequence, the Urysohn subsequence principle yields $X_n \to X_{\infty}$ in $L^1$.

Note that the implication 1 $\implies$ 4 is a special case of the generalized DCT, which holds on an arbitrary measure space. Generalized DCT is often stated for sequences converging a.e., but by similar subsequence arguments as above can also be shown to hold for sequences converging in measure.