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?
 A: If you want to go with all the details starting from the definition, do it like this : 
$$
\left| \int_S (f - g) \, d\mu \right| \le \int_S |f-g| d\mu = \sup \left\{ \left. \int_S \varphi  \, d\mu \, \right| \, 0 \le \varphi \le |f-g|, \varphi \in \mathcal L(S)\right\}
$$
where I wrote $\mathcal L(S)$ for the set of all linear combinations of characteristic functions (a characteristic function is a function which is $1$ on some measurable set and $0$ elsewhere). Since for these functions, the integral is defined as 
$$
\int_S \left( \sum_{i=1}^n a_i \mathbb 1_{A_i} \right) d\mu = \sum_{i=1}^n a_i \mu(A_i),
$$
the condition $0 \le \varphi \le |f-g|$ ensures that $a_i \neq 0 \implies \mu(A_i) = 0$, hence $\int_S \varphi d\mu = 0$ for every $\varphi$ with $0 \le \varphi \le |f-g|$. Taking the supremum over a bunch of zeros is zero.
Hope that helps,
A: 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!
A: 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 $\int_{X}f = 0$.
