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. $\square$