Proving Borel Set B: Summation of Borel Functions and Lebesgue Measure Y

hellbike
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let f_n be series of borel functions. Explain why set B = {x: \sum_n f_n(x) is not convergent} is borel set.

Proof, that if\int_R |F_n|dY \leq 1/n^2 for every n then Y(B) = 0.Y is lebesgue measure.for first part i thought that set of A={x: convergent} is borel, and B=X\A so it's also borel, but i got 0 points, so I'm wrong.

for second part - it seems quite obvious for me that for every x \neq 0
lim_n Y(f^{-1}_{n}[x])->0 and i think proving this would be enough.
I tried doing this using simple functions, but got 0 points.
 
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For the first part, the problem is likely that you didn't show why the set of points where (f_n) converges should be a Borel set. It's not completely obvious.
 
and for the second part?
 
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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