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Normally, Lebesgue integral, for positive measures, is defined in the following way. First, one defines the integral for indicator functions, and linearly extend to simple functions. Then, for a general non-negative function $f$ , it is defined as the supremum of the integral of all simple functions less than or equal $f$.

However, I find another definition in the book of Lieb and Loss, "Analysis". Let $f$ be the non-negative measurable function on a measure space $X$, let $\mu$ be the measure, and define for $t >0$, $$S_f(t) = \{x \in X : f(x) >t\},$$ and $$F_f(t) = \mu(S_f(t)) .$$

Note that $F_f(t)$ is now a Riemann integrable function. Now, the Lebesgue integral is defined as:

$$ \int_X f \, d\mu = \int_0^\infty F_f(t) \, dt,$$

where the integral on the right-hand side is the Riemann integral.

Here is the google books link to the definition.

The book gives a heuristic reason why this definition agrees with the usual definition described here in the first paragraph. Now I would like to have a rigorous proof of the equivalence.

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    It follows from Fubini's theorem.2012-08-23
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    This fact is sketched for probability measure $\mu$ [here](http://en.wikipedia.org/wiki/Expected_value#General_definition) - the last formula. I wouldn't take it as a definition, though - just as a useful fact. As a definition it seems to be too hard to work with, e.g. linearity in the argument $f$ is not that easy to see.2012-08-23
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    This formula has no reason to be true if the underlying measure isn't $\sigma$-finite.2012-08-23
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    @Siminore: Maybe it is true by Fubini. Then could you please complete the proof by proving that the Riemann integral of the Riemann-integrable functions agree with this definition? Whichever way you go, I would like a complete proof, please.2012-08-23
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    @Ahriman : Counterexample, please.2012-08-23
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    Why, are you doubting what Ahriman said?2012-08-23
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    @timur: Indeed. Else, I would be a mere Thomas.2012-08-23
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    @Ahriman: The formula is true for an arbitrary measure space (when the right hand side is interpreted as a Lebesgue integral and $F_f(t)$ is replaced by the outer measure $F_f(t) = \mu^\ast(S_f(t))$ and holds as stated if $\mu$ is complete). No need to assume $\sigma$-finiteness or anything else on $\mu$. This is proved e.g. in Fremlin's [measure theory](http://www.essex.ac.uk/maths/people/fremlin/mt.htm), volume 2, 252 O.2012-08-26
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    Thank you for the reference, I'll take a look at it.2012-08-26
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    This definition seems to be a bit more intuitive in my opinion. Thank you for shareing.2015-11-02

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