Consider f:[a,b] --> R, an integrable and bounded (ain't that implied by "integrable"?!) function and consider {a_n} a sequence that converges towards a and such that a < a_n < b (for all n). Show and rigorously justify that(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\lim_{n \rightarrow \infty} \int_{a_n}^{b} f(x)dx = \int_{a}^{b} f(x)dx[/tex]

All we found is the, imo, not very rigorous and seemingly too easy,

[tex]\int_{a}^{b} f(x)dx = \int_{a}^{a_n} f(x)dx + \int_{a_n}^{b} f(x)dx \Rightarrow \lim_{n \rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{n \rightarrow \infty} \int_{a}^{a_n} f(x)dx + \lim_{n \rightarrow \infty} \int_{a_n}^{b} f(x)dx[/tex]

[tex]\Rightarrow \int_{a}^{b} f(x)dx = 0 + \lim_{n \rightarrow \infty}\int_{a_n}^{b} f(x)dx[/tex] qed

Does anyone with more insight see how to do this more rigorously or is this the way?

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# Integral proof

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