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This is a follow-up to a previous question. I have shown that if $f$ is a continuous real-valued function on $(0, 1]$ then $$\int_{[0, 1]}|f|dm = \lim_{n \to \infty} \int_{1/n}^{1}|f(x)|dx,$$

where the integral on the left is the Lebesgue integral and that on the right is the Riemann integral.

Now I'm trying to show that if $\lim_{n \to \infty} \int_{1/n}^{1}|f(x)|dx < \infty$, then $$\int_{[0, 1]}fdm = \lim_{n \to \infty} \int_{1/n}^{1}f(x)dx.$$

I'm not sure how to do this. I can't use the monotone convergence theorem as I did for the first part, because $f$ is not necessarily positive. Could the dominated convergence theorem be used? E.g. if I defined the sequence $$f_n(x) = \begin{cases}f(x)&\text{if }x\ge\frac1n\\0&\text{otherwise}\end{cases}$$ and then took $g(x) = |f(x)|$. Then clearly $0 \leq |f_n(x)| \leq g(x)$ for all $x$, and $f_n \to f$ pointwise, and $\int_{[0, 1]}g(x) dm < \infty$ by assumption. Then we have $$\int_{[0, 1]}fdm = \int_{[0, 1]} \lim f_n dm = \lim \int_{[0, 1]} f_n dm = \lim \int_{1/n}^{1}f(x)dx,$$ since $$\int_{[0, 1]} f_n dm = \int_{[\frac{1}{n}, 1]} f_n dm = \int_{1/n}^{1}f_n(x)dx = \int_{1/n}^{1}f(x)dx$$

Does this work?

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What exactly is the question? What part of the above is unclear? –  copper.hat Oct 1 '12 at 19:10
    
Well I guess my question is, "Is what I've done correct?" Since I'm new to all this I don't really have a lot of confidence in what I've written, even though it seems to work. –  rt93 Oct 1 '12 at 19:14
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Looks good to me. Another approach would be to split $f$ into $f_+(x) = \max(f(x),0)$ and similarly $f_-(x) = \min(f(x),0)$. Then $f=f_++f_-$ and $f_+ \geq 0$, $-f_- \geq 0$, so you can use the usual tricks on each separately and then stitch them back together. –  copper.hat Oct 1 '12 at 19:48

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