showing that $g=0$ almost everywhere on $[0,1]$

Solution 1:

Maybe show the collection of sets $A$ such that $\int_A g\,d\mu = 0$ is a sigma-algebra.

Here's one approach.

Let $a$ and $b$ be elements of $[0,1]$ with $a<b$. Taking sequences of rationals $(p_n)$ and $(q_n)$ in $[0,1]$ with $\lim\limits_{n\to\infty} p_n =a$ and $\lim\limits_{n\to\infty} q_n = b$, you can show that $\int\limits_a^b g\ d\mu =\lim\limits_{n\to\infty}\ \int\limits_{p_n}^{q_n}g\ d\mu$.
If $U=\bigcup\limits_{n=1}^{\infty}(a_n,b_n)$ is an arbitrary open subset of $[0,1]$, with $\{(a_n,b_n)\}$ a pairwise disjoint collection of open intervals, you can show that $\int\limits_U g \ d\mu =\sum\limits_{n=1}^\infty \ \int\limits_{a_n}^{b_n} g\ d\mu$.
If $G=\bigcap\limits_{n=1}^\infty U_n$ is an arbitrary $G_\delta$ set, with each $U_n$ an open subset of $[0,1]$ and with $U_{n+1}\subseteq U_n$ for each $n$, you can show that $\int\limits_G g\ d\mu=\lim\limits_{n\to\infty}\ \int\limits_{U_n} g\ d\mu$.
If $E$ is an arbitrary measurable subset of $[0,1]$, then $E=G\setminus N$ for some $G_\delta$ set $G$ and some null set $N$, and $\int\limits_E g\ d\mu=\int\limits_G g\ d\mu$.
Consider the measurable sets $\{x:g(x)>0\}$ and $\{x:g(x)<0\}$.