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Let $f_1,f_2$ be two positive functions on $\Omega_1, \Omega_2 \subset R^2$ with $f_1|_{\partial \Omega_1}=f_2|_{\partial \Omega_2}=0$. For every $\lambda>0$, denote the the area of the domain enclosed by $f_{i}=\lambda$ by $A_i(\lambda)$, $i=1,2$. Assume

$A_1(\lambda) \leq A_2(\lambda)$ for all $\lambda >0$.

Is

$\int_{\Omega_1}f_1 dx \leq \int_{\Omega_2}f_2 dx?$

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It is not clear what you mean by "enclosed by". If this is just the set of points where $f_i(x)>\lambda$, then the answer is "yes", since $$\int_\Omega fdx=\int_0^\infty A(\lambda)d\lambda.$$ This is just Fubuni's theorem applied to the subgraph of $f$.

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  • $\begingroup$ No, $A(\lambda)=Area({x:f(x)>\lambda})$. $\endgroup$
    – Kostya_I
    Commented Feb 23, 2016 at 22:20
  • $\begingroup$ According to your question $A(\lambda)$ is an area. Kostya's formula is correct and it is a trick frequently used in probability theory when computing the expectation of a nonnegative random variable $X$, $$E[X]=\int_0^\infty P(X>\lambda)d\lambda.$$ $\endgroup$
    – Liviu Nicolaescu
    Commented Feb 23, 2016 at 22:23
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Yes, of course, if the word "enclosed" is understood properly. This follows from the definition of Lebesgue's integral. Lebesgue's original definition was the following: this is the limit of sums $$\sum\lambda_j\mathrm{area}\{ x:\lambda_j<f\leq \lambda_{j+1}\}.$$

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  • $\begingroup$ Thanks Alexandre. I don't see how $area\{ x: \lambda_j< f_1\leq \lambda_{j+1}\} \leq area\{ x: \lambda_j< f_2\leq \lambda_{j+1}\}$ follows from the assumptions. $\endgroup$
    – A random mathematician
    Commented Feb 23, 2016 at 22:13
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    $\begingroup$ Integrate by parts and you obtain the formula in Kostia_I answer. $\endgroup$
    – Alexandre Eremenko
    Commented Feb 24, 2016 at 5:28

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