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Suppose we know that $a.e. \lim f_{n}(x) = f(x)$ and $\mu \lim p_{n}(x) = p(x)$, where $f(x)$ and $p(x)$ are bounded functions and $x\in X \subset R$. I was wondering whether we can say the following or no - $$a.e. \lim f_{n}(x)p_{n}(x) = f(x)p(x).$$

For sure we can say the above for convergence in measure. But what about convergence almost everywhere?

I'm sorry if something stupid.

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    If this were true, then setting $f_n(x)=f(x)\equiv 1$ we would conclude that convergence in measure implies convergence almost everywhere. Does that sound plausible? – 2017-02-10
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    @ByronSchmuland Maybe I stated my question somehow vague. What I mean, is in which cases we could conclude almost everywhere convergence? (somehow, like domination of $f_{n}(x)$ over $p_{n}(x)$.). What I'm really interested in, is the case where $f(x) = 0$. – 2017-02-10

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When $f_n$ converges to zero almost everywhere, in order to conclude that $p_n f_n$ also converges to zero almost everywhere, you will probably also need to assume that the sequence $p_n$ is bounded.

Here is an example to show what can go wrong, if $p_n$ converges in measure but is not bounded.

We are working on $[0,1)$ equipped with Lebesgue measure. Let's start with the typewriter sequence: For $k\geq 0$ we define for $2^k \leq n<2^{k+1}$ the indicator functions $$t_n={\bf 1}{\left[{n-2^k\over 2^k}, {n-2^k+1\over 2^k}\right)}.$$ To paraphrase Example 4 at Terry Tao's blog, this is a sequence of indicator functions of intervals of decreasing length, marching across the unit interval over and over again. So $t_n$ converges to zero in measure, but not pointwise almost everywhere.

Now if $p_n=nt_n$, then $p_n\to 0$ in measure. If $f_n=t_n/n$, then $f_n\to 0$ pointwise, but $f_n p_n=t_n$ does not converge pointwise.

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    Thank you, very very very much. It was very helpful. – 2017-02-12
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    Glad to help out! – 2017-02-12