# Exercise on convergence in measure (Folland, Real Analysis)

This one comes from Folland, Real Analysis, Problem 33 in the section titled Modes of Convergence.

Suppose $f_n \geq 0$ and $f_n \rightarrow f$ in measure, then $\int f \leq \liminf \int f_n$.

So I notice a few things first off, that since $f_n \to f$ in measure, we can find a subsequence $f_{n_j}$ which converges pointwise almost everywhere (Theorem 2.30 in Folland), and for this subsequence we may say (by Fatou's lemma using $f_n \geq 0$) that $\int f \leq \liminf \int f_{n_j}$, but it's not necessarily true that $\liminf \int f_{n_j} \leq \liminf \int f_n$, or at least I don't see how to prove it (and in general this is not true for any sequence and subsequence, while the reverse inequality is, I think).

Any tips, hints, or solutions?

• Looking at $\lim \inf f_n$ or $\lim \sup f_n$ should help. – N. S. Oct 27 '11 at 21:38
• Assume that the conclusion is wrong. You can find a subsequence $\{f_{n_k}\}$ such that $\liminf \int f_n d\mu =\lim_{k\to +\infty}\int f_{n_k}d\mu$. Now, extract from this subsequence an almost everywhere converging subsequence, and Fatou's lemma yields a contradiction. – Davide Giraudo Oct 27 '11 at 21:43
• possible duplicate of Fatou's Lemma and Almost Sure Convergence (Pt. 2) – t.b. Oct 27 '11 at 22:33
• inf can only get smaller for a larger set so your supposition that $\liminf \int f_n \leq \liminf \int f_n_j$ is true. – user9352 Nov 3 '11 at 12:22

You can pass to a subsequence $f_{n_k}$ with $\int f_{n_k} \to \liminf \int f_n$ first.
This subsequence will also converge to $f$ in measure and ... then you already know what to do.
(Urysohn subsequence principle). Let $x_n$ be a sequence of real numbers, and let $x$ be another real number, then $\liminf x_n\geq x$ iff every subsequence $x_{n_j}$ of $x_n$ has a further subsequence $x_{n_{j_i}}$ such that $\liminf x_{n_{j_i}}\geq x$.