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I am trying to solve this problem:

If $f : \mathbb{R} \to \mathbb{R}$ is a Lebesgue measurable function and $\int_{0}^{1}{f(x)dx} = 1$ (Lebesgue integral) and $E = \{x \in [0, 1] \mid f(x) > 1\}$, then what is the value of the Lebesgue integral $\int_{E}{(f(x) - 1) dx}$?

But I haven't reached any good results

Behrooz
  • 1,000
  • I guess it will be easy for you to show that $f(x) -1 >0$ on $E$. So the integral is $\geq 0$ (and it is $=$ only if $E$ is a null set). – Fabian Feb 02 '15 at 08:08
  • I don't know what you mean by "what is the value"? If $f(x)=2\sin(\pi x)$ then the value is $2\sqrt{3}/\pi-2/3$, and if $f(x)=2x$ then the value is $1/4$. Do you want some interpretation of these values? – mickep Feb 02 '15 at 08:10
  • @ mickep. What you say suggests that the value of the integral depends on the Lebegue measure of the set $E$. – Behrooz Feb 02 '15 at 08:28

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