$f = g$, if $\int_A f\,d\mu = \int_A g\,d\mu$ Let be $(\Omega,\Sigma,\mu)$ a $\sigma$-finite measure space and $f,g:\Omega\to[-\infty,\infty]$ two quasi integrable functions, i.e. the integral of the positive or negative part is finite, so that the integral is well defined. A well known statement is if $\int_A f\,d\mu = \int_A g\,d\mu,\quad \forall A\in \Sigma$. Then $f=g$ $\mu $-almost everywhere. 
Now suppose just $\int_A f\,d\mu = \int_A g\,d\mu,\quad \forall A\in \Sigma$ such that $\mu(A)<\infty$ is true. I think then still  $f=g$ $\mu $-almost everywhere is true. Is this a well knowon result aswell? I could not find it, just the first one. 

Let $(E_n)$ increase to $\Omega$ with $\mu (E_n) <\infty$ for  each $n$. Consider the trace of $(\Omega,\Sigma,\mu)$ on $E_n$, i.e. $(E_n, \Sigma_n, \mu |_{E_n})$ where $\Sigma_n =\{E_n \cap A: A \in \Sigma\}$. Apply the first case to this to see that $f=g$ almost everywhere w.r.t.  $\mu |_{E_n}$ which means $f=g$ a.e. w.r.t. $\mu$ on $E_n$. Since this is true for each $n$ we see that $f=g$ a.e..
