Dominated Convergence and Monotone Convergence Assuming that the measure space is $\sigma$-finite, how can I show that the Dominated convergence theorem implies the monotone convergence theorem?
 A: This is a nice exercise. Since the monotone convergence theorem is often the very first theorem proved in measure theory, the only tools we can safely use from measure theory are the basic properties of measures, the definition of Lebesgue integral and the dominated convergence theorem (DCT). Once you realize this, you're almost there.
Let $(X,\mu)$ be a $\sigma$-finite measure space. Take an increasing sequence $\{f_{n}\}$ of integrable functions with $f_{n} \geq 0$. Let $f = \lim_{n} f_{n}$. We want to show $\lim_{n} \int f_{n} d\mu = \int f d\mu$. (I assume that showing the measurability of $f$ is not part of the exercise.)
If $\int f d\mu < \infty$, then $\lim_{n} \int f_{n} d\mu = \int f d\mu$ by DCT.
Suppose $\int f d\mu = \infty$. If $\lim_{n} \int f_{n} d\mu= \infty$, we're done. Suppose this limit is finite, say $B$. So for the rest of the proof, we assume
$$
\int f d\mu = \infty, \quad \lim_{n} \int f_{n} d\mu = B < \infty.
$$
Lemma: If $\mu(X) < \infty$, then  $\lim_{n} \int f_{n} d\mu = \int f d\mu$ by DCT.
Proof: Since $\int f d\mu = \infty$, by the definition of Lebesgue integral for nonnegative functions, there's a simple function $s \leq f$ such that
$$
B < \int s d\mu < \infty.
$$
The finiteness of $\mu$ is needed for the second inequality above. Let $g_{n} = \min \{f_{n}, s\}$.  Since $f_{n}$ increases to $f \geq s$, $g_{n}$ increases to $s$. So by DCT,
$$
\lim_{n} \int g_{n} d\mu = \int \lim_{n} g_{n} d\mu = \int s d\mu > B.
$$
But
$$
\lim_{n} \int g_{n} d\mu \leq \lim_{n} \int f_{n} d\mu = B,
$$
a contradiction.  QED
Back to the case $(X,\mu)$ is $\sigma$-finite.  There's an increasing sequence $\{A_{j}\}$ of measurable sets such that $\cup_{j} A_{j}=X$ and $\mu(A_{j}) < \infty$. By the lemma above, for each $j$,
$$
\int_{A{j}} f d\mu = \lim_{n} \int_{A_{j}} f_{n} d\mu  \leq \lim_{n} \int f_{n} d\mu = B.
$$
Letting $j \rightarrow \infty$,
$$
\lim_{j} \int_{A_{j}} f d\mu \leq B.
$$
Since $\int f d\mu = \infty$, by the definition of Lebesgue integral for nonnegative functions, there's a simple function
$$
s = \sum_{k=1}^{m} a_{k} 1_{C_{k}} \leq f
$$
such that
$$
B < \int s d\mu = \int \sum_{k=1}^{m} a_{k} 1_{C_{k}} d\mu = \sum_{k=1}^{m} a_{k} \mu(C_{k}).
$$
(The difference here from the proof of the lemma above is that the rightmost side may not be finite.) We have
$$
\int_{A_{j}} f d\mu \geq \int_{A_{j}} \sum_{k=1}^{m} a_{k} 1_{C_{k}} d\mu = \sum_{k=1}^{m} a_{k} \mu(C_{k} \cup A_{j}).
$$
Letting $j \rightarrow \infty$,
$$
B \geq \lim_{j} \int_{A_{j}} f d\mu \geq \lim_{j} \sum_{k=1}^{m} a_{k} \mu(C_{k}\cup A_{j}) = \sum_{k=1}^{m} a_{k} \lim_{j} \mu(C_{k} \cup A_{j}) = \sum_{k=1}^{m} a_{k} \mu(C_{k}) > B.
$$
The last equality above is valid because any measure is continuous from below. The first equality is valid because the limit of the finite sum of convergent sequences is the sum of the limits of the sequences. Now we have a contradiction.
