I'm trying to prove the following statements in Folland's book.
Let $(X,\mathcal{M},\mu)$ be a measure space. If $f_n\to f$ in measure and $g_n\to g$ in measure, then $f_n+g_n\to f+g$ in measure and $f_ng_n\to fg$ in measure (in this last case, assume $\mu(X)<\infty$).
The first one is already done, but the second one is giving me trouble.
My attempt: We can take subsequences $f_{n_k}$ and $g_{n_k}$ such that $f_{n_k}\to f$ a.e. and $g_{n_k}\to g$ a.e. In this case, we know that $f_{n_k}g_{n_k}\to fg$ a.e. Note that $\mu(X)<\infty$, so we can use Egoroff's theorem to conclude that $f_{n_k}g_{n_k}\to fg$ almost uniformly. Finally, almost uniform convergence implies convergence in measure, therefore $f_{n_k}g_{n_k}\to fg$ in measure.
The problem is that I just have convergence in measure for some subsequence of $f_ng_n$ and I need convergence in measure for the whole sequence. Also I don't know if using Egoroff's theorem is unnecessary, if this is the case, I would appreciate to see some more elementary prove.
Detail: We say that $f_n\to f$ almost uniformly if for every $\varepsilon > 0$ there is some $E\subset X$ measurable, such that $\mu(E) <\varepsilon$ and $f_n\to f$ uniformly on $E^c$.
Thank you!
Best Answer
There is something which is sometimes called the subsequence principle. It states that $x_n \to x$ holds if and only if every subsequence $(x_{n_k})_k$ admits a further subsequence $(x_{n_{k_l}})_l$, which converges to $x$. Note that the limit $x$ has to be fixed.
This holds as soon as the notion of convergence is induced by a topology.
This should help you here.
Interesting sidenote: Using this property, one can show that convergence a.e. (of measurable functions on $\Bbb{R}$) is not induced by a topology.
For the proof of the principle, note that "$\Rightarrow$" is trivial (use $(x_{n_k})_k$ itself for the "further subsequence". For "$\Leftarrow$", assume that $x_n \to x$ is false. Then (by definition) there is a neighborhood $U$ of $x$ such that for each $N\in \Bbb{N}$, there is some $n_N \geq N$ with $x_{n_N}\notin U$. Inductively, this allows to construct a subsequence $(x_{n_k})_k$ with $x_{n_k}\notin U$ for all $k$. But by assumption, there is a further subsequence $(x_{n_{k_l}})_l$ which converges to $x$. In particular, for $l$ large enough, $x_{n_{k_l}}\in U$ (because $U$ is an neighborhood of $x$), contradiction.
EDIT: Here is another possible proof: Using the subsequence principle, you can also show the following equivalence on a space of finite measure:
A sequence $(f_n)_n$ of measurable functions converges in measure to $f$ iff every subsequence $(f_{n_k})_k$ has a further subsequence $(f_{n_{k_\ell}})_\ell$ which converges pointwise a.e. to $f$.
Once you have shown this characterization, you can derive your claim as a corollary.