If $f_n \in L^p$,$f_n \to f$ in measure and $|f_n-f|<C$ for all $n$ and $x$ on finite measure. Then $f \in L^p$ and $f_n \to f$ in $L^p$

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Here is the full problem:

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How I proved it: consider $g_n=|f_n-f|$. Then $g_n\to0$ in measure and $|g_n|<C$. Thus $g_n$ is uniformly integrable as we are on finite measure and so by Vitali $g_n\to0$ in $L^p$. Now $|\|f_n\|_p-\|f\|_p|\leq\|g_n\|$ now since $g_n$ and $f_n$ is integrable we know the value $\|f\|_p$ must be finite. QED

It just seems like a very weird proof of $f \in L^p$ to me. Another one I found it that since $L^p$ is a vector space $f_1+f-f_1\in L^p$ and we are done. But this also feel "unnatural". I feel like there must be a very simple reason $f \in L^p$ besides the ones I mention.

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The reason $f=f_1+f-f_1$ is a valid and simple reason, since $$ \lvert f\rvert^p\leqslant 2^{p-1}\lvert f_1\rvert^p+2^{p-1}\lvert f_1-f\rvert^p\leqslant 2^{p-1}\lvert f_1\rvert^p+2^{p-1}C^p $$ and the right hand side is integrable since $f_1\in L^p$ and the measure space has finite measure.