If $f=g$ almost everywhere, $\int f=\int g$ Let $\Omega$ be a measurable set, and let $f: \Omega \to [0, + \infty]$ and $g : \Omega \to [0, +\infty]$ be non-negative measurable functions. Show that if $f(x) = g(x)$ for almost every $x \in \Omega$, then $\int_\Omega f = \int_\Omega g$.
I was given a very short proof that says that if $f=g$ almost everywhere then up to modifying $f$ on a measure zero set, we don't change its integral so we can suppose that $f(x) \geq g(x)$ $\forall x \in \Omega$ and then use the fact that the Lebesgue integral preserves inequality for measurable functions and do the same with $g$. However, saying that we don't change the integral of $f$ when modifying its values for a measure set is exactly what we are trying to prove right ? This argument is using A to prove A...
Is there a valid proof I can be given ? I precise that we have not seen $\int_\Omega f+g = \int_\Omega f + \int_\Omega g$ yet.

You can do this going back to first principles.
Let $\phi$ be a simple function with $0 \le \phi \le f$ and let $E = \{f = g\}$. You can check that $\phi \chi_E$ is also a simple function and that $$\int \phi = \int \phi \chi_E.$$ Since $\phi \chi_E = 0$ whenever $f \not= g$ it follows that $0 \le \phi \chi_E \le g$. Consequently by the definition of the integral of $g$ you have
$$\int \phi \le \int g.$$ Take the supremum over all such $\phi$ to conclude by the definition of the integral of $f$ that
$$\int f \le \int g.$$
Now do it again, exchanging the roles of $f$ and $g$.
