2
$\begingroup$

The precise statement is if $f$ is integrable, $g$ measurable, and $f = g$ almost everywhere, then $g$ is integrable, and the integrals coincide. I use the following definition of integrable: $f$ is integrable if there is a sequence of (simple) integrable functions $\{f_n\}$ such that $f_n \to f$ almost everywhere, and it is Cauchy in the mean.

Then $\lim_n f_n = f = g$ almost everywhere, and it seems that I can use the same sequence for both $f$ and $g$, and then there is hardly anything to do. What am I missing? Thanks in advance.

$\endgroup$
3
$\begingroup$

Since $f=g$ a.e., thus, as you said: $\lim f_n=g$.

Thus g is measurable and integrable, since it belongs to the closure of the simple functions' span. The two integrals are equal since the functions differs on a zero-measure set (call it A):

$\begin{align} \int_E f=\int_{A/E}f+\int_A f=\\ \int_{A/E}f=\int_{A/E}g=\\ \int_{A/E}g+\int_A g=\int_E g \end{align}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.