I am trying to show that if $f=g$, almost everywhere on some set X, where g is lebesgue integrable, then f is also lebesgue integrable and we have
$\int_{X}fd\mu=\int_{X}gd\mu$
I have googled this an found no proof of it, but I am sure it must be easy to prove. Can anyone show me how to prove this equation?
On
You need an additional assumption: the $\sigma$-algebra $\mathcal{S}$ on $X$ is actually $\mu$-complete. This isn't too big a concern since we could define an extension $\mu(A\triangle N)=\mu(A)$ for all $A\in\mathcal{S}$ and $N$ a subset of a $\mu$-null set.
Then we have $f$ is measurable since the the inverse image $f^{-1}(B)$ of a measurable $B\in\mathcal{B}(\mathbb{R})$ differs from $g^{-1}(B)\in\mathcal{S}$ by a subset of the $\mu$-null set $(f\neq g)$, thus measurable.
Now in the approximation $g=\lim g_n$ by simple functions (e.g. taking dyadic fraction values) $$ g_n=\sum_{j} \lambda_{j,n} 1_{A_{j,n}} $$ define the analogous $$ f_n=\sum_{j} \lambda_{j,n} 1_{A_{j,n}\triangle N_{j,n}} $$ where $N_{j,n}$ is a subset of $(f\neq g)$. Then you have $f_n\to f$ if you chose $N_{j,n}$ appropriately (exercise), and $\int f_n\,\mathrm{d}\mu = \int g_n\,\mathrm{d}\mu$, so taking limit we are done.
Let $N$ be the null set such that $\forall x \in X \setminus N (f(x)=g(x))$. Then $\int_X (f(x) - g(x)) d\mu = \int_{X \setminus N} (f(x) - g(x)) d\mu + \int_N (f(x) - g(x)) d\mu = \int_{X \setminus N} 0 d\mu + 0 =0$. Thus $\int_X (f(x) - g(x)) d\mu = 0$ and $\int_X f(x) d\mu = \int_X g(x)d\mu$ by linearity.