Let $(\Omega,\mathcal{A},\mu)$ be a finite measure space. For a map $f:\Omega \to \mathbb{R}$, define

\begin{equation} f^+(\omega) = \begin{cases} f(\omega) && \text{if} \quad f(\omega) \geqslant 0,\\ 0 && \text{otherwise,} \end{cases} \end{equation}


\begin{equation} f^-(\omega) = \begin{cases} -f(\omega) && \text{if} \quad f(\omega) < 0,\\ 0 && \text{otherwise.} \end{cases} \end{equation}

Now, if $f_n$ and $f$ are measurable such functions and if $f_n \to f$ uniformly on $\Omega$, does this also imply $f_n^+ \to f^+$ and $f^-_n \to f^-$ uniformly on $\Omega$?

For example; in case $f_n(\omega), f(\omega) \geqslant 0$, I can see that for any $\varepsilon > 0$, there must exist $N\in\mathbb{N}$ such that $n \geqslant N$ implies

\begin{equation} |f_n^+(\omega) - f^+(\omega)| = |f_n(\omega) - f(\omega)| < \varepsilon \end{equation}

by the uniform convergence of $f_n$ to $f$, but what if for example the signs are different at a given $n$ at some $\omega$?

Also, I need to show this is in an effort to prove that $f_n$ is integrable for large enough $n$ assuming $f$ is integrable. Am I on the right track? Thanks in advance!


1 Answer 1


If you want to show that $f_n$ is integrable for $n$ large enough there is an easier way. Try proving $$\int_{\Omega} |f_n|\,d\mu < \infty.$$ Since $f_n$ is measurable, this will ensure $f_n$ to be integrable. Another Hint: $$\int |f_n|\, d\mu \leq \int|f_n-f|\, d\mu + \int |f|\, d\mu$$ Now you need to apply the uniform convergence and $\mu(\Omega)<\infty$.

  • $\begingroup$ Thanks for the hints! Wouldn't it be more straightforward to use the uniform convergence to infer that for any $\varepsilon$ and large enough $n$, $|f_n|\leqslant|f_n - f| + |f| < |f| + \varepsilon$ and then integrate both sides of the inequality? Or would this be violating some rules? $\endgroup$ Oct 13, 2017 at 10:14
  • $\begingroup$ No rule violation. If you write you proof down, it is even better to do this the way you suggested. $\endgroup$ Oct 13, 2017 at 10:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .