Existence of subsequence such that integration converge

Question

Let {$f_n$} be a sequence of real-valued functions on [0,1] that is uniformly bounded.

Show that there exists a subsequence $n_j$ such that $ \int_A f_{n_j} dx $ converge for each borel subset of A contained in [0,1]

I prove for sequence of Borel set $A_i$ there exists such subsequence which converges for each set.

But for each Borel subset case, i think for each borel subset, find countable collection of half-open

interval that cover A and difference of measure of A and union of those collection less than $\epsilon$

and for each such collection i can find subsequence but... i think it's wrong :(

Davide Giraudo · Accepted Answer · 2013-10-10T15:03:00.873

2

You approach contains good ideas. If we manage to find a subsequence which works for each $C_i$ and the collections $\mathcal C:=\{C_i,i\geqslant 1\}$ is an algebra which generates the Borel $\sigma$-algebra, we would be done.

We can use a diagonal argument once we find the family $\{C_i,i\geqslant 1\}$ (which can be done considering finite unions of interval with rational endpoints).

edited Oct 10 '13 at 15:03

answered Oct 10 '13 at 14:45

Davide Giraudo

181,608

What does it means $C_i$ and $C_j$? – Arturo Oct 10 '13 at 15:01
I've edited.${}$ – Davide Giraudo Oct 10 '13 at 15:03
Then $C_i$ is half-open(or open, no matter) interval which has rational endpoint or countable union of interval? – Arturo Oct 10 '13 at 15:13
Ah, i got it. Thanks a lot :). Yor're really expert at this field – Arturo Oct 10 '13 at 15:43

Existence of subsequence such that integration converge

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