I have a question about a function that is null outside a set which have measure zero being integrable or not. In the case it's integrable, I must prove that its integral is $0$. I know about this theorem that says:
If $f[a,b]→R$ is an integrable function and $g:[a,b]→R$ is bounded, then if $g$ coincides with $f$ except in a set that has measure zero, then g is integrable and $∫g=∫f$.
Can I say that the function $g(x) = 0$ coincides with $f$ except when the measure is $0$, then if $f$ is integrable, $g$ is and its integral is $0$? Now, for $f$ to be integrable, it must be bounded in the set where the measure is $0$, right? So the answer would be that the function is integrable if $f$ is bounded, but not if it's not bounded. And it's $0$ when integrable, because of the theorem.
Am I right?