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I want to prove the following:

$$ \lim_{n \rightarrow \infty} \int_{0}^{b_{n}} f_{n}(x) dx= \int_{0}^{1} f(x) dx $$ provided that $f_{n}$ is Riemann-integrable on $[0,1]$ and $f_{n} \rightarrow f$ uniformly on $[0,1]$ and $b_{1} \leq b_{2}...\leq b_{n}\le \dots$ and $b_{n} \rightarrow 1$.

I started the proof by saying that since $f_{n} \rightarrow f$ uniformly so we can exchange the limit and the integral. But, that seems to be making it almost trivial. Am I missing something very important?

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  • $\begingroup$ You are missing a limit in the second line. Is intergability in the Riemann sense oo Lebesgue sense? Your argument does not take into account the limits of integration. $\endgroup$ Mar 20 at 23:21
  • $\begingroup$ @KaviRamaMurthy I have fixed the errors now. $\endgroup$ Mar 20 at 23:30
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    $\begingroup$ Yes. When you "exchange the limit and the integral," the integrals are all occurring on the same interval. $\endgroup$ Mar 20 at 23:49
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    $\begingroup$ Do you observe that you need to handle the integrand as well as interval of integration as $n$ tends to $\infty $? You can note that $$\int_0^{b_n}f_n(x)\,dx=\int_0^1 f_n(x) \, dx-\int_{b_n} ^{1}f_n(x)\,dx$$ the first term tends to $\int_0^1 f(x) \, dx$ (due to uniform convergence). Can you handle the second term? $\endgroup$
    – Paramanand Singh
    Mar 21 at 2:33
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Use $$\left|\int_0^{b_n}f_n-\int_{0}^{1}f\right|\leq \left|\int_0^{b_n}f_n-\int_{0}^{1}f_n\right|+\left|\int_0^{1}f_n-\int_{0}^{1}f\right|$$

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