Understanding Lebesgue Measurable Functions and Sets: A Comprehensive Guide

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Homework Statement


What does it mean for a function to be lebesgue measurable?

What does it mean for a set to be lebesgue measureable?
 
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First we define the notion of an outer measure L*, which is a very intuitive generalization of the length of an interval. The outer measure of a set S in R is the inf over all collections of intervals covering S of the total length of those intervals.

Then we define the Lebesgue measure L by restricting the domain of L* to sets that we call measurable:

A set E is said Lebesgue measurable if for any set A, we have

L*(A)=L*(AE)+L*(AE^c)

This strange looking condition is a more practical characterizations due to Carathéodory of the notion of measurability introduced by Lebesgue. In either case, the condition is there to insure that the measure L will be additive. I.e. for A, B disjoint, L(AuB)=L(A)+L(B). Actually, almost all sets are measurable, and to construct one that isn't, you must make explicit use of the axiom of choice. If we work with a set theory without the axiom of choice, all sets are Lebesgue-measurable.

And a function is measurable if the preimage of any interval is a meaurable set.
 
Surely this is defined in *any* source on measure theory. It shouldn't be the job of this forum to read the bloody book on someone's behalf.
 
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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