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I am a computer science engineer and I plan to do a postgraduate on mathematics. I plan on taking a qualifying exam so I'm preparing myself on analysis topics currently. I stumbled upon this question and I am not sure if there is something missing in order to prove this. I have been using a book on analysis from Rudin and another from Bartle which contain measure theory. If this can be proved, I really appreciate any help that I can get. Any help at all.

Let $\left(X,A,\mu\right)$ be a complete measure space. Let $f$ be a non negative integrable function over X. Let $b(t)=\left\{x\in X:f(x)\geq t\right\}$. Prove that: $$\int_Xf d\mu=\int_{(0,\infty)}b(t) d\lambda$$

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  • $\begingroup$ What is the measure $\lambda$? $\endgroup$ Aug 9, 2018 at 4:36

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$b(t)$ is a set and you cannot integrate it. I think you wanted $b(t)$ to be $\mu \{x\in X: f(x) \geq t\}$. With this change the equality is an easy consequence of Fubini's Theorem: RHS $=\int_0^{\infty} \int_X I_{\{f\geq t\}} d\mu d\lambda = \int_X \int_0^{\infty}\ I_{\{f\geq t\}} d\lambda d\mu =\int_X f\, d\mu$. [I have abbreviated $\{x\in X: f(x) \geq t\} $ to $\{f \geq t\}$].

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  • $\begingroup$ Thank you. I see what you mean. This really helped a lot. $\endgroup$
    – Juan S.
    Aug 9, 2018 at 14:51
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$\lambda$ usually means the Lebesgue measure (especially when talking to probalists).

It doesn't seem right that $b(t)$ is just a set, maybe the measure of the set with respect to $\mu$? In this case, look at it as a double integral?

Maybe start with the case when $f(x)$ is a step function, then generalize the argument by appealing to the density of simple functions in the nonnegative $L^1$ functions.

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