# Using Lebesgue's Dominated Convergence Theorem to show convergence of a sequence of random variables

Let $$\xi$$ and $$\eta$$ be random variables on a probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$ such that $$\mathbb{E}[|\xi|], \mathbb{E}[|\eta|] < + \infty$$.

I'd like to show that $$\xi$$ can be approximated by the sequence of random variables $$\{\xi_n\}$$ (defined below) in the sense that

$$\lim_{n \rightarrow \infty} \int_\Omega \xi_n \text{ d}\mu = \int_\Omega \xi \text{ d}\mu$$

where $$\xi_n = \sum_{k=1}^n b_k \mathbb{1}_{B_k}$$

where $$n\in \mathbb{N}$$, $$B_k \in \mathcal{F}$$, $$b_k \in \mathbb{R}$$ $$\forall k$$, and $$\mathbb{1}_{B_k}$$ are the corresponding indicator functions.

By Lebesgue's dominated convergence theorem, it suffices to show that $$\{\xi_n\}$$ converges to $$\xi$$ pointwise and is dominated by some integrable function $$g$$ in the sense that $$\forall n \in \mathbb{N}$$, $$\forall x\in \Omega$$, $$|\xi_n(x)| \leq g(x)$$

If this is true, then $$\eta$$ can also be approximated by a sequence of random variables $$\{\eta_n\}$$ in the above sense, with

$$\eta_n = \sum_{k=1}^n a_k \mathbb{1}_{A_k}$$

for some $$A_k \in \mathcal{F}$$, $$a_k \in \mathbb{R}$$

I'd also like to show that

$$\lim_{n \rightarrow \infty} \int_\Omega \xi_n\eta_n \text{ d}\mu = \int_\Omega \xi \eta \text{ d}\mu$$

Here my idea is to write

$$|\xi_n \eta_n - \xi\eta| = |\xi_n \eta_n - \xi\eta_n + \xi\eta_n - \xi\eta|$$ $$\leq |\eta_n(\xi_n - \xi)| + |\xi(\eta_n - \eta)|$$ by the triangle inequality. If $$|\xi_n - \xi| \rightarrow 0$$, $$|\eta_n - \eta| \rightarrow 0$$, then $$|\xi_n \eta_n - \xi\eta| \rightarrow 0$$.

However, I'm having trouble showing the remaining assumptions in Lebesgue's dominated convergence theorem, namely, that:

• $$\{\xi_n\}$$ converges to $$\xi$$ pointwise
• $$\{\xi_n\}$$ is dominated by some integrable function

It is also not clear how to show that $$\{\xi_n\eta_n\}$$ is dominated by some integrable function, since the product of two Lebesgue-integrable functions (the ones dominating $$\{\xi_n\}$$ and $$\{\eta_n\}$$ isn't necessarily Lebesgue-integrable.

Any help/advice would be greatly appreciated!

• It is not clear how your $\eta_n$ and $\xi_n$ are defined. Surely the coefficients $a_k,b_k$ and sets $A_k,B_k$ depend on $\eta$ and $\xi$ right ? Otherwise there is no reason to expect the integral to converge. Commented Oct 19, 2023 at 15:37
• @Stratossupportsthestrike that's part of the problem - I'm not entirely sure how to define $A_k$, $B_k$ and the coefficients $a_k$ and $b_k$, but it seems logical to approximate $\xi$ by simple functions $\xi_n$. Commented Oct 19, 2023 at 16:08
• actually it's not clear what you're trying to do in general. You say you want the $\xi_n$ to approximate $\xi$ but you only ask for their integral to converge to that of $\xi$, that's not the same thing at all (otherwise you could just take $\xi_n := \mathbb E[\xi]$ for all $n$). So I guess what you really want is $\xi_n \to \xi$ with respect to the $L^1$ norm, is that right ? In any case can please edit your question and clarify exactly what you're trying to prove, as it's hard to help you with the question as it is now Commented Oct 19, 2023 at 16:17
• $\|\xi_n-\xi\|\to0$ implies that $\xi_n\to\xi$ pointwise almost everywhere and is equivalent to (by definition) $\xi_n\to\xi$ in $L^1$ norm. So there is not really any difference between the two things you said Commented Oct 19, 2023 at 18:20
• This is called the simple approximation theorem and can be found in very many places. "Almost everywhere", in fact ; ) The basic idea is that you only need to think about the case $\xi\ge0$ and then you can partition the range of $\xi$ quite nicely, similarly to the standard intuition of rectangles in Riemann integration (but instead of doing this on the domain we do this on the range) and these simple functions tend to $\xi$ pointwise (a.e. - recall if $\xi\in L^1$ it is not actually a function, rather it is an a.e.-equivalence class of functions) and in norm. Commented Oct 19, 2023 at 18:25

Using @FShrike's hint:

First assume $$\xi: \Omega \rightarrow \mathbb{\bar{R}}$$ is a non-negative function: $$\xi \geq 0$$

By the Simple Approximation Theorem, there exists a sequence of real valued simple functions $$\{\xi_n\}_{n=1}^{\infty}$$ on $$\Omega$$ such that $$0 \leq \xi_1 \leq \xi_2 \leq ... \leq \xi$$ and $$\xi_n \rightarrow \xi$$ pointwise.

This sequence $$\{\xi_n\}_{n=1}^{\infty}$$ satisfies all the conditions of Lebesgue Dominated Convergence Theorem (LDCT), from which we deduce that

$$\xi_n \rightarrow \xi \text{ in the } L_1 \text{ norm.}$$

In the general case when $$\xi: \Omega \rightarrow \mathbb{\bar{R}}$$ is not necessarily non-negative on all of $$\Omega$$, we can write $$\xi$$ as $$\xi = \xi^{+} - \xi^{-}$$ where $$\xi^{+} = \max(\xi, 0)$$, $$\xi^{-} = \max(-\xi,0)$$.

Then $$\xi_n^{+} \rightarrow \xi^{+}$$ and $$\xi_n^{-} \rightarrow \xi^{-}$$ in the $$L_1$$ norm, whence there is a sequence of simple functions $$\{\xi_n^{+} - \xi_n^{-}\}_{n=1}^{\infty}$$ that converges to $$\xi$$ in the $$L_1$$ norm. $$\square$$

Exactly the same logic can be applied to show that $$\xi \eta$$ can be approximated by a sequence of simple functions (the product of two measurable functions is measurable, so we can once again apply the Simple Approximation Theorem followed by LDCT).