Question about equality of Lebesgue integrals

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Please assist me with the following. I've been thinking about it for a while, but don't know where to begin.

Let g be a bounded measurable function on a measurable set A,
and h be bounded measurable functions on a measurable set B.

Suppose that \forall c \in R,

\mu{x \in A | g(x) \geq c} = \mu{x \in B | h(x) \geq c}.

Prove that \intA g = \intB h.
 
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There was a small typo, which I just corrected. The problem should make sense now.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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