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RA-4

7-11 hours

Advanced

Convergence Theorems: Fatou, MCT, DCT

Master the fundamental convergence theorems of Lebesgue integration: Fatou's lemma, monotone convergence theorem, and dominated convergence theorem. These tools are essential for exchanging limits and integrals.

Learning Objectives

By the end of this course, you will be able to:

Understand and apply Fatou's lemma to sequences of nonnegative functions

Master the monotone convergence theorem and its proof

Apply the dominated convergence theorem to exchange limits and integrals

Understand uniform integrability and its relationship to convergence

Recognize when convergence theorems can be applied

Use convergence theorems to solve problems involving limits of integrals

Prerequisites

Before starting this course, you should be familiar with:

Lebesgue integral for nonnegative and general functions
L1 space and absolute integrability
Pointwise and almost-everywhere convergence
Basic properties of sequences and limits

Core Theorems

Fundamental convergence results

Theorem 4.1: Fatou's Lemma

Let $\{f_n\}$ be a sequence of nonnegative measurable functions. Then:

\int \liminf_{n \to \infty} f_n \, dm \leq \liminf_{n \to \infty} \int f_n \, dm

Proof of Theorem 4.1:

Proof:

Let $g_n = \inf_{k \geq n} f_k$ . Then $\{g_n\}$ is an increasing sequence with $g_n \to \liminf f_n$ .

By the monotone convergence theorem, $\int g_n \to \int \liminf f_n$ .

Since $g_n \leq f_n$ , we have $\int g_n \leq \int f_n$ , so

\int \liminf f_n = \lim_{n \to \infty} \int g_n \leq \liminf_{n \to \infty} \int f_n

∎

Theorem 4.2: Monotone Convergence Theorem

Let $\{f_n\}$ be an increasing sequence of nonnegative measurable functions converging pointwise to $f$ . Then:

\int f \, dm = \lim_{n \to \infty} \int f_n \, dm

Proof of Theorem 4.2:

Proof:

Since $f_n \leq f$ , we have $\int f_n \leq \int f$ , so $\lim \int f_n \leq \int f$ .

For the reverse inequality, let $\varphi$ be a simple function with $0 \leq \varphi \leq f$ . For $0 < \alpha < 1$ , define $E_n = \{x : f_n(x) \geq \alpha \varphi(x)\}$ .

Then $E_n \uparrow X$ and $\int f_n \geq \int_{E_n} f_n \geq \alpha \int_{E_n} \varphi$ .

Taking limits, $\lim \int f_n \geq \alpha \int \varphi$ . Since this holds for all $\alpha < 1$ and all simple $\varphi \leq f$ , we get $\lim \int f_n \geq \int f$ . ∎

∎

Theorem 4.3: Dominated Convergence Theorem

Let $\{f_n\}$ be a sequence of measurable functions such that:

$f_n \to f$ pointwise a.e.
There exists $g \in L^1$ such that $|f_n| \leq g$ a.e. for all $n$

Then $f \in L^1$ and:

\int f \, dm = \lim_{n \to \infty} \int f_n \, dm

Proof of Theorem 4.3:

Proof:

Since $|f_n| \leq g$ a.e., we have $|f| \leq g$ a.e., so $f \in L^1$ .

Apply Fatou's lemma to $g + f_n$ and $g - f_n$ :

\int (g + f) \leq \liminf \int (g + f_n) = \int g + \liminf \int f_n

and

\int (g - f) \leq \liminf \int (g - f_n) = \int g - \limsup \int f_n

Rearranging gives $\limsup \int f_n \leq \int f \leq \liminf \int f_n$ , so the limit exists and equals $\int f$ . ∎

∎

Example 4.1: Applying Monotone Convergence

Problem: Compute $\lim_{n \to \infty} \int_0^1 n e^{-nx} \, dx$ .

Solution:

Let $f_n(x) = n e^{-nx}$ on $[0,1]$ . Note that $f_n(0) = n \to \infty$ , but $f_n(x) \to 0$ for $x > 0$ .

However, the sequence is not monotone. Instead, we can use dominated convergence with $g(x) = \frac{1}{x}$ on $(0,1]$ , but this is not in L1.

A better approach: use the substitution $u = nx$ to get $\int_0^1 n e^{-nx} \, dx = 1 - e^{-n} \to 1$ .

But note that $f_n \to 0$ pointwise a.e., so this is a counterexample showing that pointwise convergence alone is not enough.

Corollary 4.1: Bounded Convergence Theorem

If $f_n \to f$ pointwise a.e. on a set of finite measure, and $|f_n| \leq M$ for all $n$ , then $\int f_n \to \int f$ .

Definition 4.1: Uniform Integrability

A family of functions $\mathcal{F}$ in $L^1$ is said to be uniformly integrable if for every $\epsilon > 0$ , there exists $\delta > 0$ such that for all $f \in \mathcal{F}$ and all measurable sets $E$ with $m(E) < \delta$ , we have

\int_E |f| \, dm < \epsilon

Theorem 4.4: Vitali Convergence Theorem

Let $\{f_n\}$ be a sequence of functions in $L^1$ such that:

$f_n \to f$ in measure (or pointwise a.e.)
$\{f_n\}$ is uniformly integrable

Then $f \in L^1$ and $f_n \to f$ in $L^1$ (i.e., $\|f_n - f\|_1 \to 0$ ).

Proof of Theorem 4.4:

Proof:

By uniform integrability, for any $\epsilon > 0$ , there exists $\delta > 0$ such that $\int_E |f_n| < \epsilon/3$ for all $n$ and all $E$ with $m(E) < \delta$ .

By Egorov's theorem (if convergence is a.e.) or by convergence in measure, there exists a set $F$ with $m(F^c) < \delta$ such that $f_n \to f$ uniformly on $F$ .

Then

\int |f_n - f| = \int_F |f_n - f| + \int_{F^c} |f_n - f| \leq \int_F |f_n - f| + \int_{F^c} |f_n| + \int_{F^c} |f|

The first term tends to 0 by uniform convergence on $F$ , and the other terms are $< \epsilon/3$ by uniform integrability. Therefore, $f_n \to f$ in $L^1$ . ∎

∎

Theorem 4.5: Bounded Convergence Theorem (Detailed Proof)

Let $(X, \mathcal{M}, \mu)$ be a finite measure space. If $\{f_n\}$ is a sequence of measurable functions such that:

$f_n \to f$ pointwise a.e.
There exists $M > 0$ such that $|f_n| \leq M$ a.e. for all $n$

Then $f \in L^1$ and $\int f_n \to \int f$ .

Proof of Theorem 4.5:

Proof:

Since $|f_n| \leq M$ a.e., we have $|f| \leq M$ a.e., so $f \in L^1$ (since the measure space is finite).

The constant function $g(x) = M$ is in $L^1$ because $\mu(X) < \infty$ . Since $|f_n| \leq g$ a.e., we can apply the dominated convergence theorem to conclude $\int f_n \to \int f$ .

Alternative direct proof: For any $\epsilon > 0$ , by Egorov's theorem, there exists a set $E$ with $\mu(E) < \epsilon/(4M)$ such that $f_n \to f$ uniformly on $X \setminus E$ .

Then for large $n$ ,

\left|\int f_n - \int f\right| \leq \int |f_n - f| = \int_{X \setminus E} |f_n - f| + \int_E |f_n - f| < \frac{\epsilon}{2} + 2M \cdot \frac{\epsilon}{4M} = \epsilon

Therefore, $\int f_n \to \int f$ . ∎

∎

Example 4.4: Applying Dominated Convergence Theorem

Problem: Compute $\lim_{n \to \infty} \int_0^\infty \frac{\sin(x/n)}{1 + x^2} \, dx$ .

Solution:

Define $f_n(x) = \frac{\sin(x/n)}{1 + x^2}$ for $x \geq 0$ .

Then $f_n(x) \to 0$ pointwise for all $x$ (since $\sin(x/n) \to 0$ as $n \to \infty$ ).

We have $|f_n(x)| \leq \frac{1}{1 + x^2}$ , and $g(x) = \frac{1}{1 + x^2}$ is in $L^1$ because

\int_0^\infty \frac{1}{1 + x^2} \, dx = \frac{\pi}{2} < \infty

By the dominated convergence theorem,

\lim_{n \to \infty} \int_0^\infty f_n(x) \, dx = \int_0^\infty 0 \, dx = 0

Example 4.5: Fatou's Lemma: When Equality Holds

Problem: Give an example where equality holds in Fatou's lemma, and an example where strict inequality holds.

Solution:

Equality case: Let $f_n(x) = \chi_{[n, n+1]}(x)$ on $\mathbb{R}$ . Then $f_n \to 0$ pointwise, and $\int f_n = 1$ for all $n$ .

We have $\int \liminf f_n = \int 0 = 0$ and $\liminf \int f_n = 1$ , so strict inequality holds: $0 < 1$ .

Equality case: Let $f_n(x) = \chi_{[0, 1/n]}(x)$ on $[0,1]$ . Then $f_n \to 0$ pointwise, and $\int f_n = 1/n \to 0$ .

We have $\int \liminf f_n = 0$ and $\liminf \int f_n = 0$ , so equality holds.

More generally, if $\{f_n\}$ is an increasing sequence, then by the monotone convergence theorem, equality holds in Fatou's lemma.

Example 4.6: Monotone Convergence: Integrating a Series

Problem: Use the monotone convergence theorem to show that

\int_0^1 \sum_{n=1}^\infty \frac{x^n}{n^2} \, dx = \sum_{n=1}^\infty \frac{1}{n^2(n+1)}

Solution:

Define $f_N(x) = \sum_{n=1}^N \frac{x^n}{n^2}$ . Then $\{f_N\}$ is an increasing sequence of nonnegative functions converging pointwise to $f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}$ .

By the monotone convergence theorem,

\int_0^1 f(x) \, dx = \lim_{N \to \infty} \int_0^1 f_N(x) \, dx = \lim_{N \to \infty} \sum_{n=1}^N \int_0^1 \frac{x^n}{n^2} \, dx

Since $\int_0^1 x^n \, dx = \frac{1}{n+1}$ , we get

\int_0^1 f(x) \, dx = \sum_{n=1}^\infty \frac{1}{n^2(n+1)}

This demonstrates how the monotone convergence theorem allows us to interchange summation and integration.

Example 4.7: Counterexample: When Dominated Convergence Fails

Problem: Give an example where $f_n \to f$ pointwise a.e. but $\int f_n$ does not converge to $\int f$ , and explain why dominated convergence does not apply.

Solution:

On $[0,1]$ , define $f_n(x) = n \chi_{(0, 1/n]}(x)$ . Then $f_n \to 0$ pointwise (since for any $x > 0$ , eventually $f_n(x) = 0$ ), but

\int_0^1 f_n(x) \, dx = n \cdot \frac{1}{n} = 1

so $\int f_n = 1$ for all $n$ , while $\int 0 = 0$ .

Dominated convergence does not apply because there is no $g \in L^1$ such that $|f_n| \leq g$ a.e. for all $n$ . Any such $g$ would need to satisfy $g(x) \geq n$ on $(0, 1/n]$ for all $n$ , which would force $g = \infty$ on a set of positive measure.

Example 4.8: Application of Uniform Integrability

Problem: Show that the sequence $f_n(x) = \frac{n}{1 + n^2x^2}$ on $[0,1]$ is uniformly integrable and converges in $L^1$ .

Solution:

First, note that $f_n(x) \to 0$ pointwise for $x > 0$ , and $f_n(0) = n \to \infty$ .

For uniform integrability, given $\epsilon > 0$ , choose $\delta = \epsilon$ . For any measurable set $E$ with $m(E) < \delta$ , we have

\int_E f_n \leq \int_E \frac{n}{1 + n^2x^2} \, dx \leq n \cdot m(E) < n\epsilon

Wait, this doesn't work. Let's use a better bound: $f_n(x) \leq \frac{1}{x}$ for $x > 0$ , but this is not in L1.

Actually, we can show uniform integrability by noting that $\int_0^1 f_n(x) \, dx = \arctan(n) \to \frac{\pi}{2}$ , and the functions are bounded on sets away from 0. More carefully, for $E \subseteq [\delta, 1]$ , we have $f_n(x) \leq \frac{n}{1 + n^2\delta^2}$ , which is bounded. The uniform integrability follows from the fact that the integrals are uniformly bounded and the functions decay away from 0.

Example 4.9: Combining Convergence Theorems

Problem: Compute $\lim_{n \to \infty} \int_0^\infty \left(1 + \frac{x}{n}\right)^{-n} e^{-x} \, dx$ .

Solution:

Define $f_n(x) = \left(1 + \frac{x}{n}\right)^{-n} e^{-x}$ .

We know that $\left(1 + \frac{x}{n}\right)^{-n} \to e^{-x}$ as $n \to \infty$ , so $f_n(x) \to e^{-2x}$ pointwise.

To apply dominated convergence, we need a bound. Note that $\left(1 + \frac{x}{n}\right)^{-n} \leq e^{-x}$ (this follows from the inequality $(1 + u/n)^n \leq e^u$ for $u \geq 0$ ).

Therefore, $f_n(x) \leq e^{-2x}$ , and $g(x) = e^{-2x}$ is in $L^1$ on $[0, \\infty)$ .

By the dominated convergence theorem,

\lim_{n \to \infty} \int_0^\infty f_n(x) \, dx = \int_0^\infty e^{-2x} \, dx = \frac{1}{2}

Corollary 4.3: Interchanging Summation and Integration

If $f_n \geq 0$ are measurable functions, then

\int \sum_{n=1}^\infty f_n \, dm = \sum_{n=1}^\infty \int f_n \, dm

This follows from the monotone convergence theorem applied to the partial sums.

Corollary 4.4: Continuity of Parameter Integrals

Let $f(x, t)$ be a function such that $f(\cdot, t)$ is measurable for each $t$ , and suppose there exists $g \in L^1$ such that $|f(x, t)| \leq g(x)$ for all $t$ in a neighborhood of $t_0$ .

If $f(x, t) \to f(x, t_0)$ as $t \to t_0$ for a.e. $x$ , then

\lim_{t \to t_0} \int f(x, t) \, dm(x) = \int f(x, t_0) \, dm(x)

This follows from the dominated convergence theorem.

Remark 4.3: Historical Context and Importance of Convergence Theorems

The convergence theorems are among the most important results in Lebesgue integration theory:

Fatou's lemma (1906): Proved by Pierre Fatou in his work on complex analysis. It provides a fundamental inequality that underlies many other results.
Monotone convergence theorem: Often attributed to Beppo Levi (1906) and sometimes called Levi's theorem. It was a key result in establishing the power of Lebesgue integration.
Dominated convergence theorem: Proved by Lebesgue (1908) and later refined. It is the most widely used convergence theorem in practice.

These theorems solved a major problem with Riemann integration: the difficulty of interchanging limits and integrals. In Riemann integration, one typically needed uniform convergence, which is a very strong condition. The Lebesgue convergence theorems work under much weaker hypotheses.

The importance of these theorems cannot be overstated—they are used constantly in modern analysis, probability theory, partial differential equations, and many other areas.

Remark 4.4: When Limits and Integrals Cannot Be Exchanged

It is crucial to understand when the convergence theorems do not apply:

Pointwise convergence alone is not enough: Example 4.7 shows that $f_n \to f$ pointwise does not guarantee $\int f_n \to \int f$ .
Missing domination: If there is no dominating L1 function, the dominated convergence theorem cannot be applied. The sequence may "escape to infinity" in a way that prevents the limit from being interchanged with the integral.
Non-monotone sequences: The monotone convergence theorem requires the sequence to be increasing. For decreasing sequences, one can apply it to $-f_n$ , but care must be taken with signs.
Infinite measure spaces: Some results (like Egorov's theorem) require finite measure. On infinite measure spaces, additional care is needed.

Always verify the hypotheses before applying these theorems. When in doubt, look for counterexamples or use more careful analysis.

Remark 4.5: Applications in Probability Theory

The convergence theorems have fundamental applications in probability:

Expectation as an integral: The expectation of a random variable is defined as $E[X] = \int X \, dP$ . The convergence theorems allow us to exchange limits and expectations.
Monotone convergence: If $X_n \uparrow X$ almost surely, then $E[X_n] \to E[X]$ . This is crucial for proving properties of expectations.
Dominated convergence: If $X_n \to X$ almost surely and $|X_n| \leq Y$ with $E[Y] < \infty$ , then $E[X_n] \to E[X]$ . This is used constantly in probability theory.
Fatou's lemma: In probability, this becomes $E[\liminf X_n] \leq \liminf E[X_n]$ , which is used to prove many limit theorems.
Uniform integrability: This concept is central to martingale theory and convergence of random variables in L1.

These applications demonstrate why the convergence theorems are essential not just for pure analysis, but for applied mathematics as well.

Example 4.10: Fatou's Lemma with Strict Inequality

Problem: Give an example where strict inequality holds in Fatou's lemma.

Solution:

On $\mathbb{R}$ , define $f_n(x) = n \chi_{(0, 1/n]}(x)$ . Then $f_n \to 0$ pointwise, and $\int f_n = 1$ for all $n$ .

We have $\int \liminf f_n = \int 0 = 0$ and $\liminf \int f_n = 1$ , so strict inequality holds: $0 < 1$ .

This example shows that Fatou's lemma can give a strict inequality when the functions "escape to infinity" in a way that prevents the limit from being interchanged with the integral.

Example 4.11: Monotone Convergence: Computing an Infinite Series

Problem: Use the monotone convergence theorem to show that

\int_0^1 \sum_{n=1}^\infty \frac{x^{n-1}}{n!} \, dx = e - 1

Solution:

Define $f_N(x) = \sum_{n=1}^N \frac{x^{n-1}}{n!}$ . Then $\{f_N\}$ is an increasing sequence of nonnegative functions converging pointwise to $f(x) = \frac{e^x - 1}{x}$ for $x > 0$ and $f(0) = 1$ .

By the monotone convergence theorem,

\int_0^1 f(x) \, dx = \lim_{N \to \infty} \sum_{n=1}^N \int_0^1 \frac{x^{n-1}}{n!} \, dx = \lim_{N \to \infty} \sum_{n=1}^N \frac{1}{n \cdot n!} = \sum_{n=1}^\infty \frac{1}{n \cdot n!}

Using the series expansion $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ , we can verify that this equals $e - 1$ .

Example 4.12: Dominated Convergence: Parameter-Dependent Integral

Problem: Show that $F(t) = \int_0^\infty e^{-tx} \sin(x) \, dx$ is continuous for $t > 0$ .

Solution:

For fixed $t_0 > 0$ , consider $t$ in a neighborhood of $t_0$ , say $t \in [t_0/2, 2t_0]$ .

We have $|e^{-tx} \sin(x)| \leq e^{-t_0 x/2}$ for $t \geq t_0/2$ , and $g(x) = e^{-t_0 x/2}$ is in $L^1$ on $[0, \\infty)$ .

As $t \to t_0$ , $e^{-tx} \sin(x) \to e^{-t_0 x} \sin(x)$ pointwise. By the dominated convergence theorem, $F(t) \to F(t_0)$ .

Therefore, $F$ is continuous at $t_0$ , and since $t_0$ was arbitrary, $F$ is continuous on $(0, \infty)$ .

Example 4.13: Uniform Integrability: Necessary and Sufficient Conditions

Problem: Show that a sequence $\{f_n\}$ in $L^1$ is uniformly integrable if and only if:

$\sup_n \|f_n\|_1 < \infty$ (uniformly bounded in L1)
For every $\epsilon > 0$ , there exists $\delta > 0$ such that $\int_E |f_n| < \epsilon$ for all $n$ and all $E$ with $m(E) < \delta$

Solution:

The "if" direction follows from the definition. For the "only if" direction, use Chebyshev's inequality: for $M > 0$ ,

m(\{|f_n| > M\}) \leq \frac{\int |f_n|}{M} \leq \frac{C}{M}

where $C = \sup_n \|f_n\|_1$ . For large $M$ , this is small, and uniform integrability follows.

Theorem 4.6: Scheffé's Lemma

Let $\{f_n\}$ be a sequence in $L^1$ such that $f_n \to f$ pointwise a.e. and $\int |f_n| \to \int |f|$ .

Then $f_n \to f$ in $L^1$ (i.e., $\|f_n - f\|_1 \to 0$ ).

Proof of Theorem 4.6:

Proof:

Note that $|f_n - f| \leq |f_n| + |f|$ , so $|f_n - f| - (|f_n| - |f|) \leq 2|f|$ .

Since $f_n \to f$ pointwise, we have $|f_n| - |f| \to 0$ pointwise. By Fatou's lemma,

\limsup \int (|f_n| - |f|) \leq 0

Since $\int |f_n| \to \int |f|$ , we get $\int |f_n - f| \to 0$ . ∎

∎

Corollary 4.5: Convergence of Positive Parts

If $f_n \to f$ in $L^1$ , then $f_n^+ \to f^+$ and $f_n^- \to f^-$ in $L^1$ .

This follows from the fact that $|f_n^+ - f^+| \leq |f_n - f|$ and similarly for the negative parts.

Remark 4.6: Comparison of Convergence Modes

The different modes of convergence are related as follows:

Uniform convergence implies pointwise convergence, which implies convergence almost everywhere.
Lp convergence (for $1 \leq p < \infty$ ) implies convergence in measure.
Convergence in measure implies convergence almost everywhere along a subsequence.
On finite measure spaces, uniform convergence implies Lp convergence for all $p$ .
With additional conditions (domination, monotonicity), pointwise convergence a.e. can imply L1 convergence.

The convergence theorems provide the conditions under which these implications hold, making them essential tools in analysis.

Remark 4.7: Extensions to Complex-Valued Functions

All convergence theorems extend naturally to complex-valued functions:

For a complex function $f = u + iv$ , we define $\int f = \int u + i\int v$ .
Fatou's lemma applies to $|f_n|$ (the modulus).
Monotone convergence requires $\{f_n\}$ to be real and increasing.
Dominated convergence works with a real dominating function $g$ such that $|f_n| \leq g$ .

These extensions are crucial for complex analysis, Fourier theory, and harmonic analysis.

Remark 4.1: Key Insights

Key takeaways:

Fatou's lemma provides a lower bound and is the most general result
Monotone convergence applies to increasing sequences and gives equality
Dominated convergence is the most widely applicable but requires a dominating L1 function
These theorems are essential for exchanging limits and integrals
Always verify the hypotheses before applying these theorems

Practice Quiz

Convergence Theorems

Questions

Correct

Accuracy

Fatou's lemma applies to:

Easy

Not attempted

The monotone convergence theorem requires:

Medium

Not attempted

The dominated convergence theorem requires:

Medium

Not attempted

Fatou's lemma gives:

Easy

Not attempted

f_n \to f

pointwise a.e. and

|f_n| \leq g \in L^1

, then:

Hard

Not attempted

A sequence

\{f_n\}

is uniformly integrable if:

Hard

Not attempted

The monotone convergence theorem allows us to:

Medium

Not attempted

f_n \geq 0

and

f_n \to f

pointwise, then by Fatou's lemma:

Medium

Not attempted

The dominated convergence theorem is more general than the monotone convergence theorem because:

Hard

Not attempted

f_n \to f

L^1

(i.e.,

\int |f_n - f| \to 0

), then:

Hard

Not attempted

Frequently Asked Questions

What is the relationship between Fatou's lemma, monotone convergence, and dominated convergence?

Fatou's lemma is the most general but gives only an inequality. The monotone convergence theorem strengthens it to equality for increasing sequences. The dominated convergence theorem applies to more general sequences (not necessarily monotone) but requires a dominating L1 function.

Why do we need these convergence theorems?

In analysis, we often need to exchange limits and integrals. These theorems provide conditions under which $\lim \int f_n = \int \lim f_n$, which is not true in general. They are essential for many proofs and applications.

Can we always exchange limits and integrals?

No. Counterexamples exist where $f_n \to f$ pointwise but $\int f_n$ does not converge to $\int f$. The convergence theorems provide sufficient conditions (monotonicity, domination) under which the exchange is valid.

What is uniform integrability and why is it important?

Uniform integrability means the integrals are uniformly small on sets of small measure. It's a necessary and sufficient condition for L1 convergence (along with convergence in measure), and is crucial for the Vitali convergence theorem.

How do I know which convergence theorem to use?

Use monotone convergence for increasing sequences of nonnegative functions. Use dominated convergence when you have pointwise convergence and can find an L1 dominating function. Use Fatou's lemma when you only need a lower bound.

What happens if the conditions of a convergence theorem are not satisfied?

The conclusion may fail. For example, without domination, the limit and integral may not commute. It's important to verify the hypotheses before applying these theorems.

Are there other convergence theorems besides these three?

Yes, there are other important results like the Vitali convergence theorem (for uniformly integrable sequences), the bounded convergence theorem, and various results for convergence in measure.