One measurable $\lim$ and one theorem

Question

How we can prove following theorem?

Let $f_n \ge 0 $ be measurable, $\lim f_n = f $ and $f_n \le f$ for each $n$. Show that $$\int f(x)dx=\lim_n \int f_n(x)dx $$

Any idea would be highly appreciated.

Answer 1

Fatou's lemma gives $\int f = \int \liminf_n f_n \le \liminf_n \int f_n$ (the values may be infinite).

If $\liminf_n \int f_n < \infty$, then $f$ is integrable and the result follows from the dominated convergence theorem.

Otherwise, we have $\lim_n \int f_n = \infty$, and since $0 \le f_n \le f$, we have $\int f_n \le \int f$ and so $\int f = \infty$.

Aside: The dominated convergence theorem states that if $|f_n| \le g$, $f_n(x) \to f(x)$ and $g$ is integrable, then $\int f_n \to \int f$.

To apply dominated convergence theorem, let $g=f$. Note that all the functions are non-negative, so $|f_n| = f_n$ and by assumption $f_n \le f = g$.

Since $\int f \le \liminf_n \int f_n < \infty$ we see that $g=f$ is integrable, hence the conditions of the theorem are satisfied and so $\lim_n \int f_n = \int f$.