One task in the book Analysis 2 by Tao is to prove the following Proposition 8.2.6:
Let $(E,\mathcal{E},\mu)$ a measure space and $f,g \colon E \to [0,\infty]$ $\mathcal{E}$-measurable functions on $E$.
If $f=g$ almost everywhere, then $\int_E f = \int_E g$.
My problem is not to prove the claim. For example let $A = \{f \neq g\} \in \mathcal{E}$, then $f = \chi_A f + \chi_{A^c}f$, $g = \chi_A g + \chi_{A^c}g$ and $\chi_{A^c}f = \chi_{A^c}g$. So we have $\int_E f = \int_A f + \int_{A^c} f$, when we know that interchanging sum and integral is possible. Then it's easy to show the rest. ($\chi$ is indicator function)
But the only properties I have available are:
- $\int_E f = \sup \left\{ \int_E g : g \text{ is simple, } 0 \leq g \leq f \right\}$
- $f_E f = 0 \Leftrightarrow f=0$ almost everywhere
- $\int_E cf = c \int_E f$, $c \in \mathbb{R}$, $c \geq 0$
- $f \leq g \Rightarrow \int_E f \leq \int_E g$
Hint
Your statement is equivalent to prove that if $N$ is a null set, then $\int_Nf=0$.
First, see what happen with simple function.
Then, if $f\geq 0$, from one of your property, there is a sequence of simple function $(g_n)$ s.t. $$\lim_{n\to \infty }\int_N g_n=\int_N f.$$
Finally, set $f^+(x)=0\vee f(x)$ and $f^-(x)=-(f(x)\wedge 0)$, where $a\wedge b$ denote the minimum of $a$ and $b$ and $a\vee b$ denote the maximum. Remark that $f^\pm \geq 0$. Then, if $f$ is unspecified, write it as $f=f^+-f^-$, and apply the previous step.