0
$\begingroup$

This is related to Lang, Real and Functional Analysis pg 142, Chpt VI, Sec 5, Cor 5.9. The following $L^1(\mu)$ means completion of $L_1$ integrable step functions space completion under $L_1$ norm. And all maps in $L^1(\mu)$ are targeted to either reals. Actually, it works for even for all finite dimensional Banach spaces.

Cor 5.9 Let $f$ be $\mu-$measurable. Then $f\in L^1(\mu)$ iff $|f|\in L^1(\mu)$. More generally, if there exists an element $g\in L^1(\mu)$ s.t. $g\geq 0$ and $|f|\leq g$, then $f\in L^1(\mu)$.

$\mu-$measurable means that $f$ is pointwise limit of sequence of steps funcions upto difference over a set of measure $0$. The proof is basically done by DCT. In general, $\mu-$measurable is not necessarily in $L^1$ due to non-integrability.

$\textbf{Q:}$ The book says above corollary explains the role of positivity in integration theory. What is the role of positivity? Why it explains it? All I could see is the following. It basically says $f\in L^1$ iff $f^+,f^-\in L^1$.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.