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}
and
\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!