Is an arbitrary measurable function that vanish locally $\mu$-almost everywhere integrable, with integral $0$?

I came across this question when studying measure theory:

Is an arbitrary measurable function that vanish locally $\mu$-almost everywhere integrable, with integral $0$?

The motivation behind this question is that I have proved that an arbitrary measurable functions that vanish almost everywhere is integrable, with measure $0$. I did this by first using the following proposition

Proposition$\quad$Let $(X,\mathscr{A},\mu)$ be a measure space, let $f$ be a $[0,+\infty]$-valued $\mathscr{A}$-measurable function on $X$, and let $\{f_n\}$ be a nondecreasing sequence of functions in $\mathscr{S}_+$ such that $f(x)=\lim_{n\to\infty}f_n(x)$ holds at each $x$ in $X$. Then $\int fd\mu = \lim_{n\to\infty}\int f_nd\mu$.

to deal with nonnegative functions and then using the decomposition $f=f^+-f^-$.

So I want to explore whether a similar result would hold for an arbitrary measurable function that vanish locally$\mu$-almost everywhere. But I have trouble proceeding a similar proof. So I am questioning whether this is correct or not?

Could someone please help me with it? If it is wrong, please provide a counterexample; if it is correct, please provide a proof. Thank you very much!

A property holds locally$\mu$-almost everywhere if the set of points at which it fails to hold is locally$\mu$-null.

A set $N$ is called locally $\mu$-null if for each set $A$ that belongs to $\mathscr{A}$ and satisfies $\mu(A)<+\infty$ the set $A\bigcap N$ is $\mu$-null.