If ${f_n}$ converges to $f$ in $L_p$ sense and to $f'$ point-wisely, does it mean $f=f' a.e.$?

Question

The question came into my mind when I read a theorem from Kubrusly's "Measure Theory: a First Course", saying that

if $f_n\rightarrow f'$ uniformly and $f_n\rightarrow f''$ in $L_p$ sense, then $f'=f''$ $\mu.a.e$.

I'm pondering whether the above statement is true if the condition of uniform convergence is weakened to point-wise convergence, that is, whether we have the following statement:

if $f_n\rightarrow f'$ point-wise and $f_n\rightarrow f''$ in $L_p$ sense, then $f'=f''$ $\mu.a.e$.

Having in mind that $f_n\rightarrow f''$ in $L_p$ does not mean that $f_n$ converges point-wisely (even a.e.), I personally feel that the proof of it may be nontrivial but so far I do not have a counter-example in my mind, either.

Thanks for any help and if this statement turns out to be rather trivial, I would say sorry for the waste of everyone's time in advance.(^_^)

Answer 1

For any sequence $f_n \to f''$ in $L^p$, there exists a subsequence such that $f_{n(k)} \to f''$ almost everywhere as $k \to \infty$. There are (at least) two possibilites to prove this statement:

• In the proof of the Riesz-Fischer theorem (which states that $L^p$ is a complete space), one usually constructs such a sequence (see e.g. René Schilling: Measures, Integrals and Martingales).
• $L^p$-convergence implies convergence in measure. Since it is well-known that convergence in measure also implies the existence of such a sequence, the claim follows.

Since on the other hand $f_{n(k)} \to f'$, it follows from the (a.e.) uniqueness of pointwise limits that $f' =f''$ almost everywhere.