# Monotone Convergence Theorem

## Theorem

Consider the measure space $(X,{\mathcal {M}},\mu )$ and suppose $\{f_{n}\}$ is a sequence of non-negative measurable functions, $f_{n}:X\to [0,+\infty ]$ such that $f_{n}\leq f_{n+1}$ for all $n\in \mathbb {N}$ . Furthermore, $\lim _{n\to +\infty }f_{n}=f(=\sup _{n}f_{n})$ . Then

$\lim _{n\to +\infty }\int f_{n}=\int \lim _{n\to +\infty }f_{n}.$ ## Proof

First we prove that $\lim _{n\rightarrow +\infty }\int f_{n}\leq \int \lim _{n\rightarrow +\infty }f_{n}$ .

Since $f_{n}\leq f_{n+1}$ for all $n\in \mathbb {N}$ , we have $f_{n}\leq \lim _{n\rightarrow +\infty }f_{n}$ and further $\int f_{n}\leq \int \lim _{n\rightarrow +\infty }f_{n}$ .

Sending $n\rightarrow +\infty$ on LHS gives us the result.

Then we only need to prove that $\lim _{n\rightarrow +\infty }\int f_{n}\geq \int \lim _{n\rightarrow +\infty }f_{n}$ . In this regard, denote $g=\lim _{n\rightarrow +\infty }f_{n}$ (which is measurable as the limit of measurable functions) and consider a simple function $\phi$ so that $0\leq \phi \leq g$ .

Now, for $n\in \mathbb {N}$ and some $t\in (0,1)$ , define the increasing sequence of measurable sets $E_{n}=\{x\in X\mid f_{n}-t\cdot \phi >0\}$ . Since $f_{n}\leq f_{n+1}$ , we have $E_{n}\subseteq E_{n+1}$ for $n\in \mathbb {N}$ . Moreover, we have $X=\cup _{n=1}^{\infty }E_{n}$ since $g=\sup _{n\in \mathbb {N} }f_{n}$ .

Recall that the set function $\nu (A)=\int _{A}\phi \,d\mu$ for $A\in {\mathcal {M}}$ is a measure on $(X,{\mathcal {M}},\mu )$ . We now see that $\lim _{n\rightarrow +\infty }\int _{X}f_{n}\geq \lim _{n\rightarrow +\infty }\int _{E_{n}}f_{n}\geq t\cdot \lim _{n\rightarrow +\infty }\int _{E_{n}}\phi =t\cdot \int _{X}\phi$ , where we have used the fact that $\nu (X)=\nu (\cup _{n=1}^{\infty }E_{n})=\lim _{n\rightarrow +\infty }\nu (E_{n})$ for the last equality. Taking the supremum over simple functions such that $0\leq \phi \leq g$ and the supremum over $t\in (0,1)$ yields the required inequality.