don't understand a step in $\int |f| d \mu < \infty $ then $f$ finite $\mu$-a.e. on $X$.

Question

Hi All: This is probably not a good question but I can't figure it out. I am reading a document on probability and there is a proposition and a proof of it which I write below:

Proposition 4.14: If $ f: X \rightarrow R $ is an integrable function, meaning that $\int |f| d \mu < \infty $ then $f$ finite $\mu$-a.e. on $X$.

Proof: We may assume that $f$ is positive without loss of generality. Suppose that $E = [ x \in X : f = \infty ] $ has non-zero measure. Then, for any $ t > 0$, we have $ f > t \chi_A $ so $ \int f d \mu \ge \int t \chi_A d \mu = t \mu(E) $ which implies that $\int f d \mu = \infty$.

My lack of understanding is where they get $f > t \chi_A$. Thanks.

Answer 1

$f >tI_E$ because $x \in E$ implies $f(x)=\infty >t =tI_E(x)$ and $ x \notin A$ implies $f(x) >0 =tI_E$. Note that the assumption that $f>0$ is can be replaced by the assumption that $f \geq 0$. In this case we get $f \geq tI_E$ by a similar argument and this is good enough to draw the conclusion.

Actually I don't see why we can assume that $f>0$. By considering $f^{+}$ and $f^{-}$ we may reduce the proof to the case $f \geq 0$ and this is good enough.

Answer 2

As GReyes pointed out, it must be $\chi_E$. You can see the inequality $f>t\chi_E$ in this way: Fix $t>0$, for $x\in E$, $f(x)=\infty$ so $f(x)>t$; for $x\notin E$, $f(x)>0=t\chi_E(x)$ by assumption. Combining these two points completes the inequality.