Define the functions $h_n^+$ and $h_n^-$ as follows: $$h_n^+ (x) = \sup \{f_m (x) \colon m \ge n\}$$ $$h_n^- (x) = \inf \{f_m (x) \colon m \ge n\}$$ These suprema and infima exist because, for every $x$ , $|f_n (x)| \le g(x)$ . These functions enjoy the following properties:

For every $n$ , $|h_n^\pm| \le g$

The sequence $h_n^+$ is decreasing and the sequence $h_n^-$ is increasing.

For every $x$ , $\lim_{n \to \infty} h_n^\pm (x) = f(x)$

Each $h_n^\pm$ is measurable.

The first property follows from immediately from the definition of supremum. The second property follows from the fact that the supremum or infimum is being taken over a larger set to define $h_n^\pm (x)$ than to define $h_m^\pm (x)$ when $n > m$ . The third property is a simple consequence of the fact that, for any sequence of real numbers, if the sequence converges, then the sequence has an upper limit and a lower limit which equal each other and equal the limit. As for the fourth statement, it means that, for every real number $y$ and every integer $n$ , the sets $$\{x \mid h_n^- (x) \ge y\} \hbox{\hskip 0.5in and \hskip 0.5in} \{x \mid h_n^+ (x) \le y\}$$ are measurable. However, by the definition of $h_n^\pm$ , these sets can be expressed as $$\bigcup_{m \le n} \{x \mid f_n (x) \le y\} \hbox{\hskip 0.5in and \hskip 0.5in} \bigcup_{m \ge n} \{x \mid f_n (x) \le y\}$$ respectively. Since each $f_n$ is assumed to be measurable, each set in either union is measurable. Since the union of a countable number of measurable sets is itself measurable, these unions are measurable, and hence the functions $h_n^\pm$ are measurable.

Because of properties 1 and 4 above and the assumption that $g$ is integrable, it follows that each $h_n^\pm$ is integrable. This conclusion and property 2 mean that the monotone convergence theorem is applicable so one can conclude that $f$ is integrable and that $$\lim_{n \to \infty} \int h_n^\pm (x) \,d\mu(x) = \int \lim_{n \to \infty} h_n^\pm (x) \,d\mu(x)$$ By property 3, the right hand side equals $\int f(x) \,d\mu(x)$ .

By construction, $h_n^- \le f_n \le h_n^+$ and hence $$\int h_n^- \le \int f_n \le \int h_n^+$$ Because the outer two terms in the above inequality tend towards the same limit as $n \to \infty$ , the middle term is squeezed into converging to the same limit. Hence $$\lim_{n \to \infty} \int f_n (x) \,d\mu(x) = \int f (x) \,d\mu(x)$$