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