# Monotone Convergence Theorem

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## Theorem

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

${\displaystyle \lim _{n\to +\infty }\int f_{n}=\int \lim _{n\to +\infty }f_{n}.}$[1]

## Proof

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

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

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

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

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

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

## References

1. Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second edition, §2.2