In the proof at (1) appears $\int_E(g-f)$. My question is: How we can be sure $g-f$ is well defined (is not $\infty-\infty$) Since Lebesgue integrability do not involve bounded. How g-f is well defined?
2026-03-30 03:38:02.1774841882
Lebesgue's Dominated Convergence Theorem $(g-f)$ is finite, is well defined?
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Both $f$ and $g$ are Lebesgue-integrable. Therefore, the set where $f$ can be $+\infty$ or $-\infty$ is a set of measure 0. The same goes for $g$.
You are right, there can be points where $g-f$ is really not defined, but because of the argument above this can only happen on a set with measure 0 which can be neglected when dealing with Lebesgue-integration.