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?

You do not need boundedness to conclude that integral of a function over a set of measure zero is zero. It is true for all measurable functions, bounded or not.
To see this, let $f$ be a function which is zero almost everywhere (i.e. except on a set of measure zero). Then let
$$ E = \{ x \in X \; | \; f(x) > 0 \}$$
$$ F = \{ x \in X \; | \; f(x) < 0 \}$$
Clearly both $E$ and $F$ are of sets of measure zero.
Now let $\phi_{n}$ be a sequence of simple functions such which increase to $f\chi_{E}$ pointwise. Then $\forall n \in \mathbb{N}, \phi_{n}$ is zero except possibly on $E$ which is of measure zero. And hence $\int_{X}\phi_{n} = 0\; \forall n$ and hence $\int f\chi_{E} = 0$ and similarly, $\int f\chi_{F} = 0$ and hence $\int_{X}f = 0$.
