Proofs of Three Fundamental Theorems in Measure Theory – Part 4: Lebesgue’s Dominated Convergence Theorem

Integral with a limit
Integral with a limit

Alright! The last one! The Dominated Convergence Theorem(DCT) gives another restraint that can lead to an exchange of $lim$ sign in expectation. As it gives a different restraint than MCT, it is also very useful as in many proofs in the measure theory, the assumption of DCT is a given. Let’s look at the statement!

Lebesgue’s Dominated Convergence Theorem: If the r.v.’s $X_n$ converge a.s. to $X$ and if $|X_n|\le Y$ a.s. and $Y \in L1$, all n, then $X_n \in L1$, $X \in L1$, and $E[X_n] \rightarrow E[X]$.

Proof: Notice that we have three conclusions. 1. $X_n \in L1$. 2. $X \in L1$. 3. $E[X_n] \rightarrow E[X]$.

The assumption gives us a sequence of random variables $X_n$ converging to $X$. Let’s construct a new random variable $U=lim inf_{n \rightarrow \infty}X_n$ and $V=lim sup_{n \rightarrow \infty}X_n$. We can see $U$ is the smallest accumulation and $V$ is the largest accumulation. But note that both are in the range of the accumulation. The accumulation can be understood as the limit. The following graph is a nice illustration to help the reader better understand(The graph shows when $X_n$ takes one input, so the y-axis should be denoted as $X_n(\omega)$, from wikipedia).

Because by assumption, we have $X_n \rightarrow X$ a.s., this means the accumulation is fixed and we have $U=V=X$ a.s. i.e. $lim sup X_n = lim X_n = lim inf X_n$.

From the assumption, we know $|X_n| \le Y$ a.s.. Because the expectation is a positive operator, we have $E[|X_n|] \le E[Y]$ and since $Y \in L1$, $E[Y]$ is bounded, therefore, $E[|X_n|]$ is also bounded for all n.

Notice that $E[|X_n|]=E[{X_n}^{+}+{X_n}^{-}]=E[{X_n}^{+}]+E[{X_n}^{-}]$ where ${X_n}^{-}=-min(X_n,0)$. As $E[|X_n|]$ is finite, we have $E[{X_n}^{+}]+E[{X_n}^{-}]$ finite, which means each of them is finite. Then $E[X_n]=E[{X_n}^{+}]-E[{X_n}^{-}]$ is also finite. Therefore, we reach conclusion 1, $E[X_n] \in L1$.

Similarly, because $|X_n| \le Y$ a.s.,when we take the limit of both sides, we have $lim_{n \rightarrow \infty}|X_n| \le lim_{n \rightarrow \infty} Y$, i.e. $|X| \le lim_{n \rightarrow \infty} Y$ since $X_n \rightarrow X$. So $|X| \le Y \in L1$. Now, apply the same trick above, we have $X \in L1$. Second conclusion reached.

Now, let’s look at the last one, and the most interesting one. Remember how Fatou’s lemma is the next best thing under exchange of $lim$? We are going to cleverly utilize Fatou’s lemma to squeeze and give the exchangeability.

First, since $|X_n| \le Y$, we have $-Y \le X_n \le Y$ a.s.. Because $-Y \le Y \in L1$, we have $-Y \in L1$. Therefore, we have $X_n \ge -Y$ and $-Y \in L1$. Remember anything familiar? That’s the prerequisite for Fatou’s lemma! Hence, we have $E[lim inf_{n \rightarrow \infty}X_n] \le lim inf_{n \rightarrow \infty} E[X_n]$, or using our notation of $U$, $E[U] \le lim inf_{n \rightarrow \infty} E[X_n]$, denoted as (i).

Similarly, from $X_n \le Y$a.s., we have $-X_n \ge -Y$ and we already showed $-Y \in L1$, so apply Fatou’s lemma again! We have another inequality: $E[lim inf_{n \rightarrow \infty}(-X_n)] \le lim inf_{n \rightarrow \infty} E[-X_n]$. Now is the tricky part, we set $V=lim sup X_n$ and we can deduce that $-V =lim inf (-X_n)$. This is clear when we look at the graph I showed you before and here I changed y-axis to $-X_n(\omega)$, just to make reader see visually. By flipping the sign, we essentially flip $lim inf$ to $lim sup$ and vice versus.

Recall that we already have $E[-V] \le lim inf_{n \rightarrow \infty}E[-X_n]=-lim sup E[X_n]$ where the last equality is applying the same trick on $E[X_n]$. By linearity of expectation, we have $E[-V]=-E[V]$. To connect this equality with the last one, we have $-E[V] \le -lim sup_{n \rightarrow \infty} E[X_n]$, changing negative sign, we have $E[V] \ge lim sup_{n \rightarrow \infty} E[X_n]$, denoted as (ii).

In the beginning, we showed $U=V=X a.s.$. Now, we applied the lemma 1 in MCT post which states that if $X=Y a.s.$, then $E[X]=E[Y]$. We have $E[U]=E[V]=E[X]$, denoted as (iii). The final key observation to connect all the dots and squeeze into an equality:

$E[X]=E[U] \le lim inf_{n \rightarrow \infty}E[X_n] \le lim sup_{n \rightarrow \infty} E[X_n] \le E[V]=E[X]$

The first and last equality is from what we just showed at (iii). The second inequality is (i), the third inequality is by the basic fact that $lim inf \le lim sup$ and the fourth inequality is from (ii).

By the squeeze theorem, as the start and the end is the same thing, the middle thing should all equal to the start and the end. Therefore, we have $lim inf_{n \rightarrow \infty} E[X_n] = lim sup_{n \rightarrow \infty} E[X_n]=E[X]=E[lim_{n \rightarrow \infty} X_n]$. The last equality holds because $X_n \rightarrow X$.

And we know if $lim inf= lim sup$, then it means both converge to $lim$. Therefore, $lim inf_{n \rightarrow \infty} E[X_n] = lim sup_{n \rightarrow \infty} E[X_n]=lim_{n \rightarrow \infty} E[X_n]$. Finally, we have $lim_{n \rightarrow \infty} E[X_n]=E[lim_{n \rightarrow \infty} X_n]$. $\Box$

This is an interesting proof! It is pretty clever in manipulating simple stuffs, like flipping signs and constructing to apply squeeze theorem. Hope you had fun!

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