# F and g are equal almost everywhere iff int(f)dm = int(g)dm for every measurable set.

1. Apr 1, 2012

### jdinatale

I feel like I got the first part, but the converse is a little tricky. I'm not sure if I am allowed to conclude at the end that f(x) must equal g(x) almost everywhere.

2. Apr 1, 2012

### jgens

Re: f and g are equal almost everywhere iff int(f)dm = int(g)dm for every measurable

Your argument for the converse does not work. Notice that since $f,g$ are measurable it follows that the sets $E_1 = \{x \in X:f(x) < g(x)\}$ and $E_2 = \{x \in X:g(x) < f(x)\}$ are measurable. Now use the fact that $\int_{E_1} f-g = 0$ and $\int_{E_2} f-g = 0$ to prove that $\mu(E_1 \cup E_2) = 0$.