# Doubt inequality almost everywhere

According to the book the following is true:

Let $f$ be a measurable function and suppose $|f| \leq h$ where $h$ is a real-valued function and $h$ is integrable. Then f is integrable.

My question is: suppose instead we have the inequality $|f| \leq h$ but almost everywhere and of course $h$ assumed to be integrable. Is it still true that f is integrable?

Would this follow because the sets of measure zero "don't count" when integrating?

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According to what book? – Mariano Suárez-Alvarez Nov 11 '10 at 19:59
@Mariano: The book. – Arturo Magidin Nov 11 '10 at 21:09
@Mariano: Fremlin's book. – student Nov 12 '10 at 1:11

Let $E$ be set of measure zero such that $|f|\leq h$ on $X-E$. Then let $g=f\chi_{X-E}$ (that is, redefine $f$ to be zero in $E$). Then $|g|\leq h$ everywhere, so $g$ is integrable. Then you have \begin{align*} \int_X g\,d\mu &= \int_{(X-E)\cup E}g\,d\mu = \int_(X-E)g\,d\mu + \int_E g\,d\mu\\ &= \int_{X-E}f\,d\mu + 0 = \int_{X-E}f\,d\mu = \int_{X-E}f\,d\mu + \int_E f\,d\mu\\ &= \int_{X}f\,d\,\mu, \end{align*} because $\int_E fd\,\mu$ exists and is equal to $0$, since $\mu(E)=0$ (so any sequence of simple functions that converges to $f$ will always have integral equal to $0$ over $E$).
@AD. Dang; it just happened again in a different post, this time with \{ and \\{. I did write in meta: meta.math.stackexchange.com/questions/1115/… – Arturo Magidin Nov 11 '10 at 6:17