$\|f_n-f\|_\infty\to 0\implies \|f_n-f\|_p\to 0$ for any $p\in [1,\infty)$. Suppose that $(X, S, \mu)$ is a measure space and that $f_n:X\to \mathbb R$ is a sequence of measurable functions converging to $f$ in $L^\infty(\mu)$. It is given that $\mu(X)<\infty$.  Then, I want to prove that $f_n\to f$ in $L^p(\mu)$ for any $p\in [1,\infty)$.
Remark: This problem has an additional hypothesis - $\|f_n\|_\infty\le M$ for all $n\in \mathbb N$.  But I think that this is not required.
I tried to prove it as follows:-
We have $\|f_n-f\|_\infty\to 0.\implies$ Given any $\epsilon>0, \exists N \left(n\ge N\implies \|f_n-f\|_\infty\lt \epsilon\right)$.
$\implies |f_n-f|\ge \epsilon $ on a set of measure $0$. (This is by definition of essential supremum $\|h\|_\infty$)
$\implies |f_n-f|\lt \epsilon$ a.e.
$\implies (\int|f_n-f|^p)^\frac 1p\lt \epsilon (\mu (X))^\frac 1p\implies \|f_n-f\|_p\to 0$.
Is this proof correct? Thanks.
 A: If $f$ or some $f_n$ do not have a finite $L^\infty$ norm, you cannot be sure that they are integrable in $L^p$. Other than that, your proof is correct.
Fixing $\epsilon>0$ and suitably large $n$: $$\|f_n-f\|_p^p=\int_{|f_n-f|>\epsilon}|f_n-f|^p+\int_{|f_n-f|\le\epsilon}|f_n-f|^p\le 0+\epsilon^p\cdot\mu(X)$$Letting $\epsilon$ vary you get what you need.
So, for all large $n$, $|f_n-f|$ is $L^p$-integrable and this norm is vanishing. However, that doesn't mean $f_n\to f$ in $L^p$ strictly speaking since it is possible that neither $(f_n)$ nor $f$ are (representatives of) elements of $L^p$. To say [something] tends to [something else] in a topological space, a fundamental starting point is that each [something] must belong to that space. The question is a little loose because it phrases: "$f_n\to f$ in $L^\infty$" but doesn't elaborate on whether or not $f_n,f$ belong to $L^\infty$ or to any $L^p$-space for that matter. I assume that the question intends only to say, $f_n,f$ are measurable functions and $f_n-f$ is (eventually) in $L^\infty$ and tends to zero therein (I explain this assumption below).
With this assumption, the case that $(\|f_n\|_\infty)_n$ is unbounded is a genuine concern. That implies: $$\|f\|_\infty=\|f-f_{n_k}+f_{n_k}\|_\infty\ge|\|f-f_{n_k}\|_\infty-\|f_{n_k}\|_\infty|\longrightarrow+\infty$$Along some subsequence $(n_k)_k$. So $f$ does not have a finite $\infty$-norm. That doesn't mean $f$ is not in $L^p$, but it leaves the possibility open. If $f$ is not in $L^p$, it makes no sense to say $(f_n)_n\to f$ in $L^p$, but it does make sense to say $|f_n-f|\to0$ in $L^p$ since we know that, eventually, $(f_n-f)\in L^p$. Similarly, if $(\|f_n\|_\infty)_n$ is bounded, we are guaranteed that all the $(f_n)_n$ and $f$ are in $L^\infty$ and in particular are in $L^p$ for every $p$ since the measure space is finite.
Then again, I might pedantically insist that when the question says: "$f_n\to f$ in $L^\infty$" that this means $f_n,f\in L^\infty$ for every $n$, in which case there is no concern. However, that makes the concern about the boundedness of $(\|f_n\|_\infty)_n$ redundant since this is impossible, making the question bad in a second way.
In conclusion, whether or not this is a real concern depends on how you read the question.
