Last updated on: Sun Jul 19 09:18:22 IST 2026

Convergence theorems

1. Monotone convergence theorem (MCT)
1.1. MCT for Riemann integration?
1.2. Why not decreasing?
1.3. Problem set 1
2. Fatou's lemma
2.1. Problem set
3. Dominated Convergence Theorem
3.1. Problem set

$\newcommand{\rightarrowL}[1]{\stackrel{L_{#1}}{\longrightarrow}}$ $\newcommand{\rightarrowA}{\stackrel{a.s.}{\longrightarrow}}$ Convergence theorems

1. Monotone convergence theorem (MCT)

We would have been very happy, had there been a result saying: Whenever $X_n\rightarrowA X$ we have $E(X_n)\rightarrow E(X)$. Unfortunately, this is not true in general (think of counterexamples). So we search for extra conditions under which it will be true.

Monotone convergence theorem (MCT) Let $X_n\rightarrowA X$. Assume

$(X_n)$ is a non-negative, non-decreasing sequence.

Then $E(X_n) \rightarrow E(X)$.

1.1. MCT for Riemann integration?

Do we have an MCT for Riemann integrals, as well? Well, no, because even if $X_n$'s are Riemann integrable, and they increase to $X,$ there is a no guaranty that $X$ will be Riemann integrable.

Here is the standard counterexample.

EXAMPLE 1: We consider $[0,1]$ equipped with the $Unif[0,1]$ probability distribution. Here density exists in a Riemann sense. Let $X$ be the Dirichlet function on $[0,1].$

Since ${\mathbb Q}\cap[0,1]$ is countable, we may enumerate it as $\{r_1,r_2,...\}.$ Take $X_n=1_{\{r_1,...,r_n\}}.$

Then the Riemann integral of each $X_n$ is $0.$ The $X_n$'s increase to $X.$ But $X$ is not Riemann integrable. ■

However, it may be shown that if the limit is a Riemann integrable, then MCT holds for Riemann integration also.

1.2. Why not decreasing?

The 'M' in MCT refers to monotone, which may mean both increasing as well as decreasing (or non-decreasing or non-incresing). However, in the theorem we require the sequence of functions to be non-increasing.

Does MCT hold for the non-increasing case? Unfortunately no, as the following counterexample shows.

EXAMPLE 2: Here we shall work with $[0,1]$ again, equipped with $Unif[0,1]$ probability distribution. Let $X_n = \frac{1}{nx}$ (set $X_n=0$ at $x=0$). Also, take $X\equi0.$ Then $\forall \omega\in[0,1]~~X_n(w)\downarrow X(w).$ But $E(X_n)=\infty$ though $E(X)=0.$

The problem is that the Lebesgue integral is defined as sup of approximations from below. ■

1.3. Problem set 1

EXERCISE 1: Give $X_n, X$ on $[0,1]$ equipped with $Unif[0,1]$ such that

$X_n\rightarrow X$ but not monotonically
$E(X_n), E(X)$ exist finitely.
But $E(X_n)\not\rightarrow E(X).$

[Hint]

Think of constant functions and step functions with just two steps.

EXERCISE 2: If $(X_n)$ is a nonincreasing sequence of nonnegative random variables converging to some random variable $X,$ and $E(X_1)<\infty,$ then show that $E(X_n)\downarrow E(X).$ What if the assumption $E(X_1)<\infty$ is dropped?

EXERCISE 3: Suppose that $X_n$'s are nonnegative random variables. Show that $$E(\sum_1^\infty X_n) = \sum_1^\infty E(X_n).$$

2. Fatou's lemma

Fatou's lemma Let $(X_n)$ be a sequence of nonnegative random variables. Then $$E(\liminf X_n) \leq \liminf E(X_n).$$

Proof: Let $Y_n = \inf\{X_k~:~k\geq n\}.$

Then, by the definition of $\liminf$, we have $Y_n\uparrow \liminf X_n.$

So, by MCT, $E(Y_n)\rightarrow E(\liminf X_n).$

Now $Y_n \leq X_n,$ and hence $E(Y_n) \leq E(X_n).$

Hence $$E(\liminf X_n) \leq \liminf E(X_n),$$ as required. [QED]

2.1. Problem set

EXERCISE 4: Consider $U\sim Unif(0,1).$ Let $X_n = U^n$ for $n\in{\mathbb N}$. Compute both sides of Fatou's lemma explicitly.

EXERCISE 5: Again let $U\sim Unif(0,1).$ Let $X_n=\left\{\begin{array}{ll}n&\text{if }U\in\left(0,\frac 1n\right)\\ 0&\text{otherwise.}\end{array}\right.$. Compute both sides of Fatou's lemma explicitly.

EXERCISE 6: Let $U\sim Expo(1).$ Let $X_n = \left\{\begin{array}{ll}e^x&\text{if }x\in(n-1,n]\\ 0&\text{otherwise.}\end{array}\right.$ for $n\in{\mathbb N}.$ Compute both sides of Fatou's lemma explicitly.

EXERCISE 7: Suppose that $(X_n)$ is a sequence of random variables with $X_n\rightarrow X.$ If $\forall n\in{\mathbb N}~~ E(X_n)\in [2,5]$, then which of the following must be true?

$E(X)\geq 2$

$E(X)\leq 5.$ Prove and/or provide counterexample(s) accordingly.

EXERCISE 8: Give examples to show that equality or strict inequality may prevail in Fatou's lemma.

EXERCISE 9: We proved Fatou's lemma from MCT. It is possible to prove MCT using Fatou's lemma quite easily. How?

EXERCISE 10: Let $(X_n)$ be nonnegative random variables such that $\forall n\in{\mathbb N}~~E(X_n) \leq M$ for some $M\in{\mathbb R}.$ If $X_n\rightarrow X,$ then show that $E(X) < \infty.$

EXERCISE 11: Let $(X_n)$ be nonnegative random variables such that $\forall n\in{\mathbb N}~~E(X_n) < \infty.$ If $X_n\rightarrow X,$ then must it be true that $E(X) < \infty?$

3. Dominated Convergence Theorem

Dominated Convergence Theorem (DCT) Let $X_n\rightarrow X.$ If $\forall n~~|X_n|\leq Y$ for some $Y$ with $E(|Y|)< \infty$, then $E|X_n-X|\rightarrow 0$ and so, in particular, $E(X_n)\rightarrow E(X).$

Proof: Clearly, $|X|\leq Y.$

So, by triangle inequality, $|X_n-X|\leq |X_n|+|X|\leq 2Y.$

Let $Z_n = 2Y-|X_n-X|.$ Then $Z_n$'s are all nonnegative random variables.

Applying Fatou's lemma to $(Z_n)$, we have $$E(\liminf Z_n)\leq \liminf E(Z_n).\hspace{1in} \mbox{(*)}$$ Now $$\liminf Z_n = 2Y-\limsup|X_n-X| = 2Y,$$ and $$\liminf E(Z_n) = 2E(Y)-\limsup E|X_n-X| .$$ So (*) becomes $$2E(Y)\leq 2E(Y)-\limsup E|X_n-X|,$$ or $\limsup E|X_n-X|\leq 0.$

Hence $E|X_n-X|\rightarrow 0,$ as required. [QED]

We can think of the conclusion of MCT or DCT as limit passing through expectation: $\lim E(X_n) = E(\lim X_n).$ This proves handy in many situations as we shall see in the problems below.

3.1. Problem set

EXERCISE 12: Let $(X_n),X$ be random variables such that $X_n\rightarrowA X$ and $\forall n\in{\mathbb N}~~|X_n|\leq Y$ for some random variable $Y$ with $E(Y) < \infty.$ Then show that $X_n\rightarrowL 1 X.$

EXERCISE 13: Let $X$ be any random variable. Show that $f(t) = E(sin(tX))$ is a continuous function.

Hint:

Enough to show that if $a_n\rightarrow a,$ then $f(a_n)\rightarrow f(a).$

EXERCISE 14: Let $X$ be any random variable. Show that $f(t) = E(\sin(t+X))$ is a differentiable function, with $f'(t) = E(\cos(t+X)).$

Hint:

Enough to show that if $a_n\rightarrow a,$ then $Y_n = \frac{f(a_n)-f(a)}{a_n-a}\rightarrow E(\cos(t+X)).$ Use the mean value theorem and boundedness of $\cos.$