Suppose that $f, (f_n)$ are nonnegative measurable functions, that $f_n \to f$ pointwise, and that $f_n \leq f$ for all n. Prove that:

$\int f = \lim_{n \to \infty} \int f_n$

My attempt

One direction seems fairly obvious. Since $f_n \leq f$ for all n, then: $\int f_n \leq \int f$ for all n.

So we should have: $\lim_{n \to \infty} \int f_n \leq \int f$

In the other direction, use Fatou’s Lemma to see that:

$\int f \leq \lim_{n \to \infty} \inf \int f_n$

However, it’s not actually clear that $\lim_{n \to \infty} \int f_n$ is well-defined, so it doesn’t necessarily make sense to get there from the $\lim_{n \to \infty} \inf$.

As a concept, my idea would then (or maybe instead?) create a subsequence from $(f_n)$ that is monotone and then invoke MCT? But I am not sure how to go about this.


Fatou's Lemma gives \begin{align*} \int f\leq\liminf\int f_{n}. \end{align*} From \begin{align*} f_{n}\leq f, \end{align*} we get \begin{align*} \int f_{n}\leq\int f, \end{align*} and hence \begin{align*} \limsup\int f_{n}\leq\int f. \end{align*} We conclude that \begin{align*} \int f\leq\liminf\int f_{n}\leq\limsup\int f_{n}\leq\int f, \end{align*} so the limit exists and \begin{align*} \lim\int f_{n}=\int f. \end{align*}

  • $\begingroup$ Can you please explain how you got the fourth line (with the limsup)? $\endgroup$ Apr 10 '18 at 20:14
  • $\begingroup$ You can take $\limsup$ both sides along with the comparison. $\endgroup$
    – user284331
    Apr 10 '18 at 20:29

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