Help with a Royden exercise of measure

Question

I'm solving the exercise 12, of section 4 The General Lebesgue Integral from the Royden's book Real Analysis 3rd edition:

Let $g$ be an integrable function on a set $E$ and suppose that $(f_n)$ is a sequence of measurable functions such that $\vert f_n(x)\vert \leq g(x)$ a.e. on $E$. Then $$\int_E \underline{\lim}f_n \leq \underline{\lim}\int_E f_n$$

I've been able to prove it using Fatou's lemma, but my math teacher asked to prove it with dominated convergence theorem, and I can't do it. Is it possible?

I'll appreciate any suggestions, thanks!

Answer 1

Since $\inf_{i\geq n} f_i \leq f_i$ for each $i \geq n$, $$\int_E \inf_{i\geq n} f_i \leq \int_E f_i$$ for each $i \geq n$ and thus $$\int_E\inf_{i \geq n} f_i \leq \inf_{i \geq n} \int_E f_i.$$

This is essentially the proof of Fatou's Lemma, but instead of dealing with non-negative functions and applying the MCT to interchange limits, you're using the fact that your sequence is dominated by $g$, which allows you to use the DCT to switch the order of the limits.

Answer 2

Let $g_n(x)=\inf_{m\geq n}f_m(x)$. Notice that $g_n$ is an increasing sequence of functions that converges to $\varliminf f_n(x)$. Moreover $|g_n|\leq g$ for all $n$, so Dominated Convergence yields: $$ \int_E \lim_{n\to\infty}g_n=\lim_{n\to\infty}\int_E g_n. $$ Whenever $\lim$ exists, we have $\lim =\varliminf$. Therefore $$ \int_E \varliminf f_n=\varliminf\int_E g_n\leq \varliminf\int_E f_n, $$ since $g_n\leq f_n$.