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The descirption on title might be confusing, but I am doing Bass's Real Analysis for Graduate Student, excercise 11.3. The problem is ask us to prove $$\int_{-\infty}^\infty |f(x)| \, dx = \int_0^\infty m( \{x:|f(x)| \geq t\}) \, dt$$

I don't know how to understand the integral on the right-hand-side. Anyone can give a hint on how to approach? Thanks!

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    The absolute value sign seems a pointless complication since what is to be proved is that if $g$ is any everywhere non-negative function then $$ \int_{-\infty}^\infty g(x)\,dx = \int_0^\infty m\left( \left\{ x: g(x)\ge t \right\} \right) \, dt. $$ – 2017-01-28
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    Sure I know, it doesn't make things any easier though. – 2017-01-28

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Hint: write $$\int_{-\infty}^\infty |f(x)|\,dx = \int_{-\infty}^\infty \int_{-\infty}^\infty 1_{0 \leq t \leq |f(x)|}\,dt\,dx.$$

What happens to the right hand side if you apply the Fubini-Tonelli theorem?

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    You've got $t$ going from $-\infty$ to $\infty$, but it ought to go from $0$ to $\infty. \qquad$ – 2017-01-28
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    @MichaelHardy $-\infty$ to $\infty$ is correct as written because the indicator in the integrand says $0 \leq t$. – 2017-01-29
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Suppose $g$ is an everywhere non-negative function. A simple function is a function $s$ whose set of values $\{s(x) : -\inftyfinite. Then the Lebesgue integral is by definition $$ \int_{-\infty}^\infty g(x) \,dx = \sup\left\{ \int_{-\infty}^\infty s(x)\,dx : s \text{ is a non-negative simple function }\le g \right\}. $$ Each of these integrals of a simple function is a sum of finitely many terms. Suppose the positive values of a simple function $s$ are $0

Note that $t \mapsto m(\{x: s(x) \ge t\})$ is a simple function, and the measure of the set of points where it is equal to $m(\{x:s(x)\ge s_k\})$ is $s_k-s_{k-1}.$