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I'm working through some problems in a first course on Measure Spaces.

The problem is:

Let $(X,\Sigma, \mu)$ be a measure space and let $f:X \to[0,\infty]$ be a measurable function such that $\int_Xfd\mu\lt\infty$. Show that $\mu\left(\{x\in X : f(x)=\infty\}\right)=0$.

The solution goes as:

Let $E=\{x\in X: f(x)=\infty\}$. Then E is measurable. Suppose $\mu(E)\gt0$. Then for each $n \in \Bbb{N}$,$$f\ge f\chi_E \ge n\chi_E .\qquad(1)$$ So,$$\infty\gt\int_Xfd\mu\ge \int_X n\chi_E d\mu=n\mu(E).\qquad $$.

But $$\lim_{n \to \infty}n\mu(E)=\infty$$ which is a contradiction. Thus $\mu(E)=0$

My question is:

i) I can't understand how the step denoted by (1) is made, particularly the reasoning behind $f\chi_E\ge n\chi_E$. Would someone please enlighten me as to how?

Many thanks in advance.

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    $\begingroup$ From the definition of $E$, restricted on $E$, what's the value of $f$? $\endgroup$ Nov 11, 2013 at 17:55

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$E$ is defined as the set on which $f(x)= \infty$. Do you agree that $\infty > n$ for all $n \in \mathbb{N}$? There's not much more to it. Multiplying $f$ by the characteristic function of $E$ is what allows you to get a function that is identical to $f$ on $E$ but vanishes everywhere else.

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