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A proof of Monotone Convergence Theorem:

Suppose an increasing sequence $f_{n}\rightarrow f$ a.e., then it is immediate that $\lim \int f_{n}\leq \int f$. The theorem claims the reverse inequality holds as well. To show this, it's sufficient to show that

$\lim \int f_{n}\geq \int \phi$

for a simple function $\phi$ with $\phi \leq f$ (since, by definition, $\int f=\sup _{\phi }\int \phi$.)

To this end, fix $0<\alpha <1$, we consider the sets

$E_{n}=\{f(x)\geq \phi (x)\}.$

Since ${f_{n}}$ is increasing, $\cup E_{n}=X$. Also, for all n

$\int f_{n}\geq \int _{E_{n}}f_{n}\geq \alpha \int _{E_{n}}\phi .$

Now we notice that $\phi dx$ is a positive measure on X. Thus, by the so-called "continuity from below" property of measures,

$\lim _{n}\int _{E_{n}}\phi =\int \phi .$

Together with the previous estimate, this implies

$\lim \int f_{n}\geq \alpha \int \phi .$

By virtue of $\phi$ being simple, taking the supremum of $\alpha$ gives

$\lim \int f_{n}\geq \int \phi .$

Taking the supremum over $\phi$ proves the claim.

Notice the argument hinges on how one partitions the range (as oppose to the domain in the Riemann case) of the functions into suitable parts, passes back to the preimages, and then applies the machinery of measure theory.

The assumption that the sequence $\{f_{n}\}$ be increasing is needed, for the areas of $\{f_{n}\}$ might "escape to infinity" and equality may fail. In the absence of this assumption, taking the infimum and applying the monotone convergence theorem yields the Fatou's lemma:

$\int \lim \inf f_{n}=\lim _{k\rightarrow \infty }\int \inf _{k\geq n}f_{k}\leq \lim _{k\rightarrow \infty }\inf _{k\geq n}\int f_{k}.$

We note that, assuming Fatou's lemma, the monotone convergence theorem follows trivially:

$\int \sup f_{n}=\int \lim \inf f_{n}\leq \lim _{k\rightarrow \infty }\inf _{k\geq n}\int f_{k}=\sup \int f_{n}.$