Let $M = (X, \mathcal{F}, \mu)$ be a D1158: Measure space.
Let $f : X \to [- \infty, \infty]$ be an D1921: Absolutely integrable function on $M$ such that

(i)	$f =_{a.e.} 0$

Result R1022: Partition of basic function into positive and negative parts provides the decomposition $f = f^+ - f^-$. Since $f$ is almost everywhere zero, one has the almost everywhere equality $f^+ = f^-$. Both $f^+$ and $f^-$ are unsigned functions $X \to [0, \infty]$, so result R1901: Unsigned integral of almost everywhere equal functions states that their integrals coincide \begin{equation} \int_X f^+ \, d \mu = \int_X f^- \, d \mu \end{equation} Now, by definition of a signed integral, $\int_X f \, d \mu : = \int_X f^+ \, d \mu - \int_X f^- \, d \mu = 0$, which proves the claim. $\square$