Lebesgue integration: $f = g$ a.e. $ \Rightarrow \int_\Omega f = \int_\Omega g$ Let $f,g : \Omega \subseteq \mathbb R^n \rightarrow [0,+\infty]$ be measurable functions with $f(x) = g(x)$ a.e. . Then I have to show that $\int_\Omega  f = \int_\Omega g$.
I may not assume that $\int_\Omega (f+g) = \int_\Omega f + \int_\Omega g$. 
This task is from Tao Proposition 19.2.6.
 A: The definition of integration that I am using:
The predicate $CSM(p)$ is a shorthand for $p$ having a countable range and $p$ is a measurable function. Let $p:X\rightarrow [0,\infty]$ be a  function such that $CSM(p)$. We will define the integral of $p$ to be:
$$\int_X p\ d\mu=\sum_{a\in p(X)}a\mu(p^{-1}(\{a\}))$$ 
Now let $f:X\rightarrow [0,\infty]$ be any measurable function, we define the upper and lower integrals of $f$ as:
$$\int_{X}^{*}f\ d\mu=\inf\{\int_{X} p \ d\mu|p:X\rightarrow[0,\infty],CSM(p),\mu(\{x\in X|f(x)>p(x)\}=0\}$$
$$\int_{*X}^{}f\ d\mu=\sup\{\int_{X} q \ d\mu|q:X\rightarrow[0,\infty],CSM(q),\mu(\{x\in X|f(x)<q(x)\}=0\}$$
Finally, we say that the integral of $f$ exists iff $\int_{X}^{*} f\ d\mu=\int_{*X}^{}f\ d\mu$ and we denote it $\int_X f\ d\mu$.
To prove your theorem 
Verify that if $f=g$ $\mu.a.e$ then:
1) For all measurable functions $p$ , $f\leq p$ a.e. iff $g\leq p$ a.e.
2) For all measurable functions $q$ , $f\geq q$ a.e. iff $g\geq q$ a.e.
Finally deduce that the upper and lower integrals of $f,g$ are equal. 
A: You can show that the set $$ \{ \int_\Omega s : 0\leq s\leq f\}=\{\int_\Omega s : 0\leq s\leq g\}$$ by using a double inclusion argument and the fact stated in the comments of the original question that simple functions which equal almost everywhere have the same integral.
Therefore their supremums are equal which respectively equal the integrals you want.
The sets I mentioned above are equivalent to what Tao calls "simple functions that minorize $f$."
