# 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?

#### Solutions Collecting From Web of "Exercise on convergence in measure (Folland, Real Analysis)"

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.

You can use the Urysohn subsequence principle, but it should be modified a little bit.

(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 $\liminf x_{n_{j_i}}\geq x$.