Problem related to signed measure

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Homework Statement


(X,S,u) a measure space and f is in L1.
Show that for any e>0, there exists a set E with u(E)<+infinity such that
| \int_{E} fdu - \int_{X} fdu |&lt;e


The Attempt at a Solution


we can define a function
v(A)=\int_{A}fdu
It is a well known result that v(A) is in fact a signed measure.

We can somehow use the property of signed measures to show that there always exist a E such that |v(E)-v(X)|<e?
 
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Don't you need sigma-finiteness here or something of the sort?
 
The question is as is, there is no mention of sigma finiteness. My attempt at the solution could be completely wrong.
 
Let's suppose it is sigma finite? then what?
 
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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