$f \in L^{1}(X)$ and $g:X \to \mathbb{R}$ measurable, such $f=g$ a.e. Prove that $g \in L^{1}(X)$ and $\int_{X} f d \mu = \int_{X} g d \mu $.

Question

Let $f \in L^{1}(X)$ and $g:X \to \mathbb{R}$ measurable, such $f=g$ a.e. Prove that $g \in L^{1}(X)$ and $\int_{X} f d \mu = \int_{X} g d \mu $.

I already proved the following lemma:

Lemma

If $f,g:X \to \mathbb{R}$ are integrable. we have that $f=g$ a.e if and only if $\int_{A} f d \mu = \int_{A} g d \mu$ for every $A\in \mathcal{M}_{X}$ where $\mathcal{M}_{X}$ is the corresponding sigma algebra.

I dont know if I can use this lemma to prove the first statement, since what I need to prove is that

$\int |g| d \mu< +\infty$.

My attempt of proof goes as follows:

If $f=g$ a.e, then $|f|=|g|$ a.e. So $\int |g| d \mu=\int |f| d \mu< \infty$ as $f \in L^{1}$ so g \in L^{1}(X).

I got the feeling im kind of lost proving this. Any help completing the proof will be apreciated

Answer 1

Maybe you can introduce the measurable function $h:X\to \mathbb{R}$ such that $h(x)=f(x)-g(x)$ for all $x\in X$, and deduce that $h=0$ a.e. in $X$. Then show that $h$ is integrable with $\int_X h\,dx=0$, and use the additivity of Lebesgue integral that $$\int_X g\,dx=\int_X (f-h)\,dx=\int_X f\,dx-\int_Xh\,dx=\int_X f\,dx.$$

Answer 2

Since $f = g$ a.e. $\{f\neq g\}$ is a null set, so partition $X = \{x\in X:f(x)=g(x)\} \cup \{x\in X : f(x)\neq g(x)\}$. Since the integral over a null-set is zero you may conclude via your lemma.