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No. My problem was exactly why ##\|f(\chi(A_n))(x)-f(\chi(A))(x)\|<\varepsilon## without further conditions on ##f##.
I am not sure. I suppose we need ##f## to be continuous, but maybe Lebesgue integrable will do. I just don't see it. We have that ##\|\chi(A_n)(x)-\chi(A)(x)\|## is small, but why does applying ##f## keep this condition?benorin said:Well we know that f is integrable and ##A_n## is measurable, is that not enough?