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Given that $\sqrt{x}f_n(x) \rightarrow g$ uniformly where each $f_n$ is integrable.

I would like to show $\lim\limits_{n\rightarrow \infty}\int_0^1 f_n(x)dx$ exists.

So I have tried to show that $f_n$ converges uniformly. But I have not succeeded, is there something else I could try?

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    @AD. yeah (I tried to co$n$vey that w$i$th the dx )but I'd be i$n$terested in looking at Lebesgue as well I just haven't thought about it yet.2011-10-18

2 Answers 2

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Try this:

$|\int_0^1 f_n(x) dx - \int_0^1 f_m(x) dx | \leq \int_0^1 \frac{1}{\sqrt{x}}|\sqrt{x}f_n(x) - \sqrt{x}f_m(x)| dx$

and use the fact that $\sqrt{x}f_n(x) \to g(x)$ uniformly to get that the sequence is Cauchy.

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    @robjohn, Sorry, I misread your post, the 6 and 3 both looked like and$s$in small font. And I misspoke above, I really just meant that the improper Riemann integral exists, so it's not really that interesting of an example.2011-10-25
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Since $\sqrt{x}f_n(x)$ converges uniformly to $g(x)$, we have that that for any $\epsilon>0$, there is an $N>0$ so that for $m,n\ge N$, $|\sqrt{x}(f_m(x)-f_n(x))|\le\epsilon$ for all $x\in[0,1]$. Thus, $|f_n(x)|\le|f_m(x)|+\frac{\epsilon}{\sqrt{x}}$. Thus, for $n\ge N$, $f_n$ is dominated by the sum of two integrable functions. Since $f_n(x)\to \frac{g(x)}{\sqrt{x}}$ pointwise, we have that $\displaystyle\lim_{n\to\infty}\int_0^1 f_n(x)\;\mathrm{d}x$ exists by Dominated Convergence.

Furthermore, $\displaystyle\lim_{n\to\infty}\int_0^1 f_n(x)\;\mathrm{d}x=\int_0^1 \frac{g(x)}{\sqrt{x}}\mathrm{d}x$ and $\displaystyle\left\|\frac{g(x)}{\sqrt{x}}\right\|_{L^1[0,1]}\le\left\|f_m\right\|_{L^1[0,1]}+2\epsilon$

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    This would be for the Lebesgue integral.2011-10-21