Two functions agreeing except on set of measure zero 
Let $f,g:S\rightarrow\mathbb{R}$; assume $f$ and $g$ are integrable over $S$. Show that if $f$ and $g$ agree except on a set of measure zero, then $\int_Sf=\int_Sg$.

Since $f$ and $g$ are integrable over $S$, we have $f-g$ also integrable over $S$. So $(f-g)(x)=0$ except for a set of measure zero. If $(f-g)$ were bounded, we could choose a partition so that the volume covering the points $x$ such that $(f-g)(x)\neq 0$ is less than any $\epsilon$, which will imply that $\int_S(f-g)=0$, and so $\int_S f=\int_S g$. But here we don't have boundedness. How can we go from here?

The integral of any measurable function over any set of measure zero is equal to zero. You needn't worry about boundedness in your question because the definition of "Riemann integrability" entails boundedness and thus your argument is complete. Let me know if you need further clarification on this point and I'm happy to discuss it with you further!
