I consider $(f_n)_{n\in \mathbb N} \in L^p(\mu)$ and $\vert|f_n-f\vert|_p \rightarrow 0 $ for $ n \rightarrow \infty$
So I can conclude: Let $ \epsilon_k =2^{-k}$
$ \forall k \ \exists n_k: \mu(\{|f_n -f_m|>\epsilon_k \}) \leq \epsilon_k \ \forall m,n \geq n_k $
Because of the $L^p$ convergence the set has to be a null set.
I cannot imagine how you can state that in a mathematical way?
$\mu (|f_n-f_m| > \epsilon_k) \leq \frac 1 {\epsilon_k^{p}} \int |f_n-f_m|^{p} < \epsilon_k$ if $\int |f_n-f_m|^{p} <\epsilon_k^{p+1}$ and this is true for $n$ and $m$ are sufficiently large. So the answer is YES. [ Note that $\int |f_n-f_m|^{p} \to 0$ because $|f_n-f_m|^{p} \leq 2^{p} (|f_n-f|^{p}+|f-f_m|^{p})$].