Let $(X,M,\mu)$ be a measure space and $1 \leq p < \infty$ fixed. Let $\{f_n\}_{n=1}^\infty$ be a sequence of functions in $L^p$, and suppose that there's $f \in L^p$ such that $f_n \to f$ in $L^p$.
Show that if $f_n(X) \subset \{1,2, \ldots, 101\}$ for all $n$, then $f(x) \in \{1,2,\ldots,101\}$ for almost every $x \in X$.
What I did is to use the fact that if $f_n \to f$ in $L^p$, then $f_n \to f$ in measure, and then there's a subsequence $\{f_{n_k}\}_{k=1}^\infty$ which converges pointwise to $f$ a.e. Let $x \in X$ such that $f_{n_k}(x) \to f(x)$. There exists $k_0$ such that $|f_{n_{k_0}}(x)-f(x)|<1$. But, since $f_{n_{k_0}}(x) \in \{1,2,\ldots,101\}$, we must have that $f(x) \in \{1,2,\ldots,101\}$.
I'd like to know if my proof is correct.
A priori, the fact that $|f_{n_{k_0}}(x)-f(x)|<1$ and $f_{n_{k_0}}$ takes only integer values between $1$ and $101$ does not guarantee that $f(x)$ is also an integer between $1$ and $101$ (we could a priori have $f(x)=0.5$ for example). However, we can proceed as follows: for $1\leqslant i\leqslant 101$, $i$ integer, let $I_i(x)=\{k\mid f_{n_k}(x)=i\}$. Since $(f_{n_k}(x))$ converges, exactly one of the $I_i(x)$ is infinite, say $I_{i(x)}$ and it forces $f(x)=i(x)\in\{1,\dots,101\}$.