$f,g:[a,b]\to\mathbb{R}$ are integrable and $X=\{x\in [a,b]: f(x)\neq g(x)\}$ has measure zero. Show that
$\int_a^bf(x)dx=\int_a^bg(x)dx.$
I proved that each set of measure zero has empty interior. So $X\subset \cup_{i=1}^n I_i$ where each $I_i$ is an open interval and $\sum_{i=1}^n|I_i|<\epsilon$, for some $\epsilon.$
Now I am wondering what to do next? I am allowed to write the integral as
$$\int_{[a,b]}=\int_{X}+\int_{[a,b]\backslash X}?$$
If yes how to show that $\int_Xf-g=0$
Use the result that if $f$ is integrable and that if $A$ is a measurable set with measure $0$ then $\int_Af(x)dx =0$. This can be proven by first showing it true for simple functions and then non-negative measurable functions by using the result that non-negative measurable functions can be represented as the increasing limit of non-negative simple functions and then finally for general functions by considering their positive and negative parts separately.
Now using this result
$$ \begin{align} \int_a^bf(x)dx &=\int_Xf(x)dx \; + \; \int_{[a,b] \cap X^c}f(x)dx \\ &= 0 \;+\;\int_{[a,b] \cap X^c}g(x)dx \\ &= \int_Xg(x)dx \; + \; \int_{[a,b] \cap X^c}g(x)dx \\ &= \int_a^b g(x)dx \end{align}$$
Where above we used the fact that $[a,b] \cap X^c=\{x \in [a,b]:f(x)=g(x)\}$.