Monotone Convergence Theorem
Statement: Suppose we have a sequence of non-negative functions $f_n$ that converges almost everywhere to $f$ on some set $D$ on a measure space. Then $\lim \int_D f_n d\mu = \int_D f d\mu$.
Proof: Since the sequence of functions is increasing, any simple function dominated by an $f_n$ is dominated by $f$. This immediately gets us $\lim \int_D f_n d\mu \leq \int_D f d\mu$.
Now, suppose $\phi$ is a simple function dominated by $f$. Pick any $0 < \alpha < 1$ and define $A_i = \{ x \in D : f_i(x) \geq \alpha \phi(x) \}$. Since the $f_n$’s are increasing, the sets $A_n$ are increasing. Also, $f(x) = 0$ or $\alpha \phi(x) < f(x)$, and since $f_n(x) \to f(x)$, it follows that each $x$ is in some $A_i$. Hence $\cup A_i = D$.
We therefore have $\int_D f_n d\mu \geq \int_{A_n} f_n d\mu$ since the functions are non-negative. On the one hand, $\int_{A_n} f_n d\mu \geq \alpha \int_{A_n} \phi d\mu$. On the other hand, define $\nu(X) = \int_X \phi d\mu$; then $\nu$ is a measure, hence since $A_n$ is increasing, we have $\lim \nu(A_n) = \nu(\lim A_n) = \nu(D) = \int_D \phi d\mu$. Thus,
$$\lim \int_D f_n d\mu \geq \alpha \int_D \phi d\mu.$$
Since $\alpha < 1$ was arbitrary, it follows that $\lim \int_D f_n d\mu \geq \int_D \phi d\mu$. Since $\phi$ was any simple function dominated by $f$, it therefore follows that $\lim \int_D f_n d\mu \geq \int_D f d\mu$. The theorem follows.
Fatou’s Lemma
Statement: Let $f_n$ be non-negative as before. Then $\int_D \liminf f_n d\mu \leq \liminf \int_D f_n d\mu$.
Proof: For each x and N, $\inf_{n \geq N} f_n(x)$ forms an increasing sequence in N. Thus by the Monotone Convergence Theorem above,
$$\int_D \displaystyle \lim_{N\to\infty} \displaystyle \inf_{n > N} f_n d\mu =
\lim_{N\to\infty} \int_D \displaystyle \displaystyle \inf_{n > N} f_n d\mu \leq
\lim_{N\to\infty} \int_D \displaystyle \displaystyle f_N d\mu =
\liminf_{N\to\infty} \int_D \displaystyle \displaystyle f_N d\mu.$$
Reverse Fatou’s Lemma
Statement: If $f_n$ is nonnegative and bounded by an integrable function $g$, then $\limsup \int f_n \leq \int \limsup f_n$.
Proof: $\liminf (-f_n) = – \limsup f_n$. By Fatou’s lemma, $\int \liminf g – f_n \leq\liminf \int g – f_n$. Since $g$ is integrable, $\int g$ can be pulled out of both sides and all the terms cancel, leaving the desired theorem.
Dominated Convergence Theorem
Statement: Suppose $f_n$ convergece to $f$ pointwise and $|f_n| \leq g$ for some integrable $g$. Then $\lim \int f_n = \int f$.
Proof: Clearly $|\lim \int f_n – f| \leq \lim \int |f_n – f|$, where $|f_n – f|$ converges to 0 and is non-negative, so it suffices to show the theorem under the assumptions that $f = 0$ and $f_n \geq 0$.
But then existence of g means we can use the reverse Fatou’s lemma to obtain
$$
0 \leq \liminf \int f_n \leq \limsup \int f_n \leq \int \limsup f_n = \int 0 = 0.
$$
Thus these are all equal and $\lim \int f_n = \int f$.