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If I have a general measure space $(X,\mu)$, and I have a sequence $f_n \to f$ in measure, and $\lvert f_n(x) \rvert \leq M$ for all $n \in \mathbb{N}$, and $x \in X$, is it true that we must also have $\lvert f(x) \rvert \leq M$ for all $x \in X$?

I believe this is true almost-everywhere: for any fixed $\varepsilon > 0$, $\mu(\{x : \lvert f\rvert > M + \varepsilon\}) \leq \mu(\{x: \lvert f_n \rvert > M\}) + \mu(\{x: \lvert f_n - f \rvert > \varepsilon\})$, and by convergence in measure of $f_n$ this means $\mu(\{x : \lvert f \rvert > M + \varepsilon\}) = 0$, so $\lvert f\rvert \leq M$ $\mu$-a.e.

But does this need hold everywhere?

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    $\begingroup$ It does not need to hold everywhere. Simply note that if $f_n = f $ a.e., then $f_n \to f $ in measure. But if e.g. $f_n=0$, but only $f=0$ a.e. (not everywhere), you can't say anything. $\endgroup$
    – PhoemueX
    Commented Mar 21, 2016 at 11:48

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If your limit is pointwise, i.e. $$ \lim_{n\rightarrow\infty}f_{n}(x)=f(x)\text{ for all }x, $$ then the bound is true for the limiting function independent of measure since $|\cdot|$ is continuous: $$ M\geq\lim_{n\rightarrow\infty}|f_{n}(x)|=|\lim_{n\rightarrow\infty}f_{n}(x)|=|f(x)|. $$ This also assumes that the objects we are working with are functions, and not equivalence classes of functions defined almost everywhere.

However, if your limit is in measure, the claim is obviously not true. Consider $f_{n}(x)=0$ for all $x$. Clearly, this converges to $f(x)=0$, whose corresponding equivalence class also contains $$ \tilde{f}(x)=\begin{cases} 0 & \text{if }x\neq0;\\ 1 & \text{if }x=0. \end{cases} $$

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