$f = g$ a.e. implies $\int_{A}f\,d\mu = \int_{A}g\,d\mu$

Question

Suppose $f \in L(X, M , \mu)$ and $g$ is a $M$-measurable real-valued function such that $f(x) = g(x)$ $\mu$-almost everywhere on $X$. I want to show that $g \in L(X, M, \mu)$ and for all $A \in M$ $$ \int_{A}f\,d\mu = \int_{A}g\,d\mu $$

I know $f$ is integrable iff $\int|f|$ is finite. Same for $g$. $E=\{x:f(x)=g(x),x\in X\}$.

Then $\int |g|d\mu= \int_E |g|d\mu +\int_{E^c} |g|d\mu = \int_E |f|d\mu < +\infty$.

So $g$ is integrable? But I am not sure how go on to show the second part?

Answer 1

You have practically already solved it. Recall that one possible definiton of $\int_Ef$ is the supremum of aproximating $R(f,E)$ the graph of $f$ under over $E$. In particular, this lets us see that if $\mu(E) = 0$ then $\int_E f = 0$.

Thus, since

$$ X = \{f = g\} \stackrel{d}\bigcup \{f \neq g\}$$

we have that, noting $E = \{f=g\}$,

$$ \int_Af \ d\mu = \int_X\chi_Af \ d\mu = \int_E\chi_Af \ d\mu + \int_{E^c}\chi_Af \ d\mu = \int_E\chi_Af \ d\mu = \int_E\chi_Ag \ d\mu = \\ = \int_E\chi_Ag \ d\mu + \int_{E^c}\chi_Ag \ d\mu = \int_X\chi_Ag \ d\mu = \int_Ag \ d\mu. $$

Answer 2

$$\int_Af\;d\mu=\int_{A\cap E}f\;d\mu+\int_{A\cap E^{\complement}}f\;d\mu=\int_{A\cap E}f\;d\mu+0=$$$$\int_{A\cap E}g\;d\mu+0=\int_{A\cap E}g\;d\mu+\int_{A\cap E^{\complement}}g\;d\mu=\int_Ag\;d\mu$$

Answer 3

Lebesgue integral is additive. This means, if $f$ is integrable on $A$, it is integrable on any measurable subset of $A$. Furthermore, if $A = \bigsqcup\limits_{k=1}^{n}A_k$, where all $A_k$ are measurable, $$\int\limits_{A}f d\mu = \sum\limits_{k=1}^{n}\int\limits_{A_k}f d\mu$$ Also, any function is integrable on a set of measure zero, and the integral equals zero. Denote the subset of $A$ where $f \ne g$ as $D$. By condition, $\mu(D) = 0$. This means $$\int\limits_{D}f d\mu = 0 = \int\limits_{D}g d\mu$$. Applying the above: $$ \int\limits_{A}f d\mu = \int\limits_{A \setminus D}f d\mu + \int\limits_{D}f d\mu = \int\limits_{A \setminus D}g d\mu + \int\limits_{D} g d\mu = \int\limits_{A}g d\mu $$