Convergence in $L^p$ of product space implies convergence in each space? Reading a paper by EM Stein (On limits of sequences of Operators, Ann of Math, 1961), the author proves that a certain sequence of functions $F_n(x, t)$, where $x$ belongs to a probability space $(X, \mu)$ and $t \in [0,1]$, is Cauchy in the space $L^p (X \times [0,1])$ endowed with the product measure and with $1 \leq p \leq 2$.
He then proceeds to claim that it is possible to find a subsequence $F_{n_k}(x, t)$ such that for (Lebesgue) almost every $t$, $F_{n_k}(x, t)$ converges in $L^p (X)$. I wonder if this is a general property of convergent sequences in the product space and, in this case, if someone more experienced can point me towards a proof, or if he probably used characteristic features of the functions $F_n$ in question and I must find a way to reconstruct his line of thought.
For the record, the functions in question are $$F_n(x,t) = \sum_{k=1}^{n} r_k (t) f_k (x) $$
where each $f_k \in L^p (X)$, their sequence satisfies $\sum \| f_k \|_{p}^p < \infty$ and $r_k$ is the $k$-th Rademacher function.
 A: Is that the thing where he shows that almost everywhere convergence implies a maximal function inequality? Cool.
It's true in general, by more or less the same argument as shows that a sequence convergent in $L^p(\mu)$ has a subsequence convergent almost everywhere. Ok, maybe we need $\sigma$-finiteness for Fubini.
Informal notation: I'm going to write $dx$ and $dy$ for the measures on $X$ and $Y$; this is going to be enough typing as it is.
If $f\in L^p(X\times Y)$ define $F:X\to L^p(Y)$ by $$F(x)(y)=f(x,y).$$In order I think to make the formulas easier to read, write $$|F(x)|=\left(\int_Y|F(x)(y)|^p\,dy\right)^{1/p}.$$ And I'm going to write $$||F||_p=\left(\int_X|F(x)|^p\,dx\right)^{1/p};$$note that $$||F||_p=||f||_p.$$
And note that $$||F+G||_p=||f+g||_p\le||f||_p+||g||_p=||F||_p+||G||_p.$$
Say $f_n\in L^p(X\times Y)$ converges in $L^p$ to $f$. Replacing $(f_n)$ by a subsequence, again to save typing, we may assume that $$\sum||f_{n+1}-f_n||_p<\infty.$$
So $$\sum||F_{n+1}-F_n||_p<\infty,$$hence $$\left|\left|\sum|F_{n+1}(x)-F_n(x)|\right|\right|_p\le\sum||F_{n+1}-F_n||_p<\infty.$$Hence $$\sum|F_{n+1}(x)-F_n(x)|<\infty$$for almost every $x$. This implies that $(F_n(x))$ converges in $L^p(Y)$ for almost every $x$.
If you don't believe it try rewriting everything without all the abbreviations...
