Lebesgue’s Monotone convergence theorem

After all the praises Lebesgue’s Monotone convergence theorem got in the last post, it’s now finally time to present it, prove it, and reap it’s benefits! Let’s do it!

The Theorem

Suppose that for every n the function


is defined and measurable. Moreover, suppose that for every x\in X:

0\leq f_1(x)\leq f_2(x)\leq f_3(x)\leq\dots

Then the function:

f(x)=\sup f_n(x)=\lim_{n\to\infty}f_n(x)

Is also defined and measurable. In addition (and that’s the important part):


Some Distinctions

Before I’ll go into the proof, let’s understand better what this theorem states.

First, the fact that for every x\in X:

0\leq f_1(x)\leq f_2(x)\leq f_3(x)\leq\dots

Is just another way to state that: The sequnce of functions \{f_n\}_{n=1}^\infty is monotonically increasing – that’s probably why the theorem is called like that.

Now for a bit more tricky part. The theorem states that the function:

f(x)=\sup f_n(x)=\lim_{n\to\infty}f_n(x)

Is well defined and measurable. However, there is even another statement inside of it. Who said that \lim_{n\to\infty}f_n(x) even exists? And even if it is, why it is the same as \sup f_n(x)?

Well, this follows from the simple fact that the sequnce of functions \{f_n\}_{n=1}^\infty is monotonically increasing, Therefore, the limit exists (it can be infinty though) and it has to be the supremum beacuse of this fact (Make sure that you understand why the limit of an increasing sequence is it’s supremum. Think about it and convince yourself!).

Moreover, even though it’s not explicitly mentioned, it’s not hard to notice that the limit is a limit of the function at a specific point. Thus, we are dealing here with a pointwise convergence!

Now for the last part of the theorem, the most important one, who states that:


Notice what actually happens here: Since f(x)=\lim_{n\to\infty} f_n(x) this is equivalent to:

\int_X\lim_{n\to\infty} f_nd\mu=\lim_{n\to\infty}\int_Xf_nd\mu

We are interchanging the order of the limit and the integral. The monotone convergence theorem basically says that this is allowed, unlike reimann integral that has no idea how to deal with pointwise convergence!

Ok, now that we understand it better, we are ready for the proof!

The proof

I’ve already explained why the first part of the theorem is true. We only need to show why we can interchange the limit and the integral.

I am going to show it in two parts – First I’ll show that:


And then I’ll show that:


And that would be enough to conclude the equality and complete the proof!

Part 1 – “limit of integral is smaller than integral of limit”

This is the easy part. Since:

f_n(x)\leq f(x)

We know that:

\int_X f_n d\mu\leq\int_Xfd\mu

Now this is an inequality of two real number, since limit preserves (weak) inequalities we get that:

\lim_{n\to\infty}\int_X f_n d\mu\leq\lim_{n\to\infty}\int_Xfd\mu=\int_Xfd\mu

As we wanted!

Part 2 – “integral of limit is smaller than limit of integral”

We want to prove that:


Which is, by definition, equivalent to:

\sup_{0\leq\varphi\leq f}\int_X\varphi d\mu\leq\lim_{n\to\infty}\int_Xf_nd\mu

However, if we will be able to show that for every 0\leq\varphi\leq f:

\int_X\varphi d\mu\leq\lim_{n\to\infty}\int_Xf_nd\mu

It would imply that the ‘limit of the integral’ is an upper bound of the set \{\int_X\varphi d\mu | 0\leq\varphi\leq f\}, thus, it is greater or equals the set’s supremum – which is exacly what we want!

So that is what I am planning to do – instead of bounding the supremum of the set, I’ll bound an arbitrary element of it. Now there are two cases:

Case 1 – \int_X\varphi d\mu=\infty

Then we will have to show that:


as well.

Using our doubtful decision to define a\cdot \mu(E)=0 if a=0 (even though \mu(E) can be \infty), we conclude that the only way the integral \int_X\varphi d\mu will be infinity, is that there is a measurable set A with \mu(A)=\infty and, \varphi(x)\equiv c>0 for every x\in A (Recall that \varphi is simple).

Ok, now, since \varphi(x)\leq f(x) for every x\in A we can conclude that:

f(x)\geq\varphi(x)\geq c

Now I want to pause for a moment in order to prove a useful substatement:

Quick substatement

If E_1\subseteq E_2\subseteq E_3\subseteq \dots are measurable sets, then:

\mu(\bigcup_{n=1}^\infty E_n)=\lim_{n\to\infty}\mu(E_n)

The proof is really easy, We can define the sequence of sets:

F_1=E_1\\F_n = E_n\setminus E_{n-1}

(Note that \bigcup_{k=1}^nF_k=E_n) Since E_1\subseteq E_2 \subseteq E_3\subseteq \dots those sets are pairwise dijoint sets and:

\bigcup_{n=1}^\infty E_n=\bigcup_{n=1}^\infty F_n

Therefore, we can use the \sigma-additivity of the measure to get:

\mu(\bigcup_{n=1}^\infty E_n)=\mu(\bigcup_{n=1}^\infty F_n)=\sum_{n=1}^\infty\mu(F_n)=\lim_{n\to\infty}\sum_{k=1}^n\mu(F_k)
\lim_{n\to\infty}\mu(\bigcup_{k=1}^n F_k)=\lim_{n\to\infty}\mu(E_n)
Back to the proof

I am now going to define the following sequence of sets:

E_n=\{x\in A:f_n(x)>\frac{c}{2}\}

Since the sequene \{f_n\} are increasing, we conclude that:

E_1\sube E_2\sube E_3\sube\dots

and use the substatement to get:

\mu(\bigcup_{n=1}^\infty E_n)=\lim_{n\to\infty}\mu(E_n)

But what is \bigcup_{n=1}^\infty E_n? This is exactly the set A! Why?

By definition E_n\sub A for every n. Thus, \bigcup_{n=1}^\infty E_n\sube A.

On the other hand, if x\in A, then f(x)=c. Since f_n\to f, there exists n large enough such that f_n(x)>\frac{n}{2}. Then x\in E_n\sube \bigcup_{n=1}^\infty E_n. Hence A\subseteq \bigcup_{n=1}^\infty E_n.

In total we have A=\bigcup_{n=1}^\infty E_n.

Great, we can conclude:

\int_Xf_nd\mu\geq\int_{E_n}f_nd\mu\overset{\text{def. of } E_n}{>}\int_{E_n}\frac{c}{2}d\mu=\frac{c}{2}\cdot\mu(E_n)



As we wanted!

Case 2 – \int_X\varphi d\mu<\infty

This one is a bit trickier, but the idea is the really similar to the first case.

Remember, we want to prove that

\lim_{n\to\infty}\int_Xf_nd\mu\geq \int_X\varphi d\mu

Ok, let’s define the set:

A=\{x\in X:\varphi(x)>0\}

Since \int_X\varphi(x) d\mu<\infty then \mu(A) has to be smaller than \infty as well, and by it’s defintion:

\int_X\varphi d\mu=\int_A \varphi d\mu

Now, let 0< \varepsilon < 1 be some positive real number, and define:

A_n=\{x\in A:f_n(x)>(1-\varepsilon)\varphi(x)\}

Once again, since the sequence \{f_n(x)\}_{n=1}^\infty is increasing, we conclude that:

A_1\sube A_2\sube A_3\sube\dots

Moreover, A=\bigcup_{n=1}^\infty A_n. Why?

By definition A_n\sube A, thus \bigcup_{n=1}^\infty A_n\sube A.

On the other hand, If x\in A, then:

\lim_{n\to\infty}f_n(x)= f(x)\geq\varphi(x)\overset{\varphi(x)\neq0, \varepsilon<1}{>}(1-\varepsilon)\varphi(x)

And f_n\to f. Then for every x\in A there exists some n\in\mathbb{N} such that:


Thus, x\in A_n\sub \bigcup_{n=1}^\infty A_n which implies A\sube \bigcup_{n=1}^\infty A_n.

So we now have A=\bigcup_{n=1}^\infty A_n and we can use the substatement to get:

\mu(A)=\mu(\bigcup_{n=1}^\infty A_n)=\lim_{n\to\infty}\mu(A_n)\\ \Downarrow \\

Moreover, note that A=(A\setminus A_n)\cup A_n is a disjoint union, so we can use the additivity of the measure to get:

\mu(A)=\mu((A\setminus A_n))\cup A_n)=\mu(A\setminus A_n)+\mu(A_n) \\ \Downarrow\\
\mu(A\setminus A_n)=\mu(A)-\mu(A_n)


\lim_{n\to\infty}\mu(A\setminus A_n)=\lim_{n\to\infty}(\mu(A)-\mu(A_n))=0

So there exists n\in\mathbb{N} large enough such that:

\mu(A\setminus A_n)<\varepsilon

For that specific n we can conclude:

\lim_{n\to\infty}\int_Xf_nd\mu\geq\int_{A_n}f_nd\mu\overset{\text{def. of } A_n}{\geq}
\int_{A_n}(1-\varepsilon)\varphi d\mu

Since \int_{A_n}(1-\varepsilon)\varphi d\mu is an integral of a simple function and A=(A\setminus A_n)\cup A_n we know that:

=\int_{A}(1-\varepsilon)\varphi d\mu=\int_{A\setminus A_n}(1-\varepsilon)\varphi d\mu+\int_{A_n}(1-\varepsilon)\varphi d\mu \\ \Downarrow \\
\int_{A_n}(1-\varepsilon)\varphi d\mu=\int_{A}(1-\varepsilon)\varphi d\mu-\int_{A\setminus A_n}(1-\varepsilon)\varphi d\mu

In total, we have:

\int_{A_n}(1-\varepsilon)\varphi d\mu=\int_{A}(1-\varepsilon)\varphi d\mu-\int_{A\setminus A_n}(1-\varepsilon)\varphi d\mu
=(1-\varepsilon)(\int_{A}\varphi d\mu-\int_{A\setminus A_n}\varphi d\mu)

Since \varphi is simple, then it is bounded. I’ll denote a bound of it by M. Now we have:

(1-\varepsilon)(\int_{A}\varphi d\mu-\int_{A\setminus A_n}\varphi d\mu)\overset{\int_X=\int_A}{=}(1-\varepsilon)(\int_{X}\varphi d\mu-\int_{A\setminus A_n}\varphi d\mu)
\overset{\varphi\leq M}{\geq}(1-\varepsilon)(\int_{X}\varphi d\mu-\int_{A\setminus A_n}Md\mu)=
(1-\varepsilon)(\int_{X}\varphi d\mu-M\cdot\overbrace{\mu(A\setminus A_n)}^{\leq\varepsilon})
\geq(1-\varepsilon)(\int_{X}\varphi d\mu-M\cdot\varepsilon)\underset{\varepsilon\to 0}{\to}\int_{X}\varphi d\mu

So that was a long process, but if you kept track with it, you know that we have just showed that:

\lim_{n\to\infty}\int_Xf_nd\mu\geq\int_{X}\varphi d\mu

And that’s exactly what we wanted!

Summary of the proof

After we finished proing both cases we can now conlcude that:


And in the first part we’ve proved that:




And the proof is done!


This proof might be a little bit complicated when you first see it – but it makes sense. It shows us how the term of a measure was the ‘barrier’ of reimann integral that prevnted it from dealing with pointwise convergence.

However,what if we want to deal with monotonically increasing sequences? what if the limit doesn’t exits?

Well, it turns out that there is a weaker version of the monotone convergence theorem which is called fatou’s lemma and it will be quite useful in some cases.

Of course, I still want to show what we can conclude from this theorem and convince you that it is a really important one.

However, this post has gotten kind of long, So all the answers – in the next one.

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