- #1

hellbike

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

Proof, that if[tex]\int_R |F_n|dY \leq 1/n^2[/tex] 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 [tex]x \neq 0[/tex]

[tex]lim_n Y(f^{-1}_{n}[x])->0[/tex] and i think proving this would be enough.

I tried doing this using simple functions, but got 0 points.

Proof, that if[tex]\int_R |F_n|dY \leq 1/n^2[/tex] 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 [tex]x \neq 0[/tex]

[tex]lim_n Y(f^{-1}_{n}[x])->0[/tex] and i think proving this would be enough.

I tried doing this using simple functions, but got 0 points.

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