# Show that $\int_{\mathbb{R}}f = \lim_{n \to \infty} \int_{\mathbb{R}}f_n$ given specific assumptions

I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help with the following problem:

If $f, f_n: \mathbb{R} \to [0, +\infty)$ measurable, $f_n \to f$ pointwise on $\mathbb{R}$ and $f_n \leq f, \forall n \in \mathbb{N}$, show that $\int_{\mathbb{R}}f = \lim_{n \to \infty} \int_{\mathbb{R}}f_n$.

The assumptions are (I always rewrite the problem to see if I understand it correctly):

1. $f$ and $f_n$ are nonnegative measurable functions.
2. $\lim_{n \to \infty}f_n = f$ (pointwise limit).
3. $f_n \leq f$ for all $n$.

If I could commute the limit and the integral then I would be done (but it is never that simple, unfortunately).

This problem looks like an application of Fatou's lemma, i.e.

If $f_n$ is a sequence of nonnegative functions, then $$\int \Big(\liminf_{n \to \infty} \ f_n \Big) \leq \liminf_{n \to \infty}\int f_n$$

but I'm having difficulities applying it to this specific problem.

• I'd say that $g_n(x) = \sup_{m \le n} f_m(x)$ is ready for the monotone convergence theorem – reuns Apr 14 '16 at 12:11
• @user1952009 since there exists a monotone subsequence that converges subsequence, monotone convergence theorem; then by the convergence of $f_n$, you can finish from there. – Andres Mejia Apr 14 '16 at 12:14

Your assumption about Fatou's lemma is correct. Note that, using your assumptions, the left hand sides equals $\int f$. Now, on the other hand, by monotonicity of the integral, $\int f_n\le \int f$ for every $n$. Can you complete the reasoning from here?
• To ensure my reasoning is correct, when you say "the left hand side equals $\int f$" (in Fatou's lemma I presume) is this because $\lim_{n \to \infty}f_n = f \implies \liminf_{n \to \infty}f_n = f$? That is, the limit is "matching" with the $\liminf$, is that correct? I also understand that, because by assumption $f_n \leq f$ for all $n$, by monotonicity of the integral $\int f_n\le \int f$ for all $n$ but it is still unclear to me as to how to complete the reasoning from here. – Von Kar Apr 14 '16 at 12:30
• @VonKar if the limit exists then the lim inf exists and equals the limit, yes (in the sense the convergence is meant, pointwise in this case). And I was referring to Fatous lemma, yes. Write down the definition of the limes inferior, then this should show you that the inequality $\int f_n \le \int f$ implies that the reverse inequality (compared to what Fatous lemma tells you) also has to hold (actually you get $\limsup \int f_n \le \int f$). – Thomas Apr 14 '16 at 12:38
• The definition I am familiar for the limit inferior is $$\liminf_{n \to \infty} \ f_n = \lim_{n \to \infty} \big(\inf_{m \geq n}f_m\big)$$ but I don't see how this implies the reverse inequality $\limsup \int f_n \le \int f$. Given the above result, the way to conclude would be since $\liminf \int f_n \geq \int f$ and $\limsup \int f_n \le \int f$ then $\lim \int f_n = \int f$. Is that correct? The part that I need to understand is the reverse inequality. – Von Kar Apr 14 '16 at 14:02
• @VonKar if $\int f_n \le \int f$ for every $n$ then $\limsup \int f_n \le \int f_n$, because of course $\sup_{m>n} \int f_m$ is then also bounded by $\int f$ from above. Of course you need to apply the definition of $\limsup$ for this. From Fatou you get $\int f \le \liminf f_n$, from the previous reasoning $\limsup \int f_n\le \int f$. Since the reverse inequality is always true you necessarily have equality. – Thomas Apr 14 '16 at 14:17