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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
\lim_{n \rightarrow \infty} \int_{a_n}^{b} f(x)dx = \int_{a}^{b} f(x)dx
All we found is the, imo, not very rigorous and seemingly too easy,
\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
\Rightarrow \int_{a}^{b} f(x)dx = 0 + \lim_{n \rightarrow \infty}\int_{a_n}^{b} f(x)dx qed
Does anyone with more insight see how to do this more rigorously or is this the way?
\lim_{n \rightarrow \infty} \int_{a_n}^{b} f(x)dx = \int_{a}^{b} f(x)dx
All we found is the, imo, not very rigorous and seemingly too easy,
\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
\Rightarrow \int_{a}^{b} f(x)dx = 0 + \lim_{n \rightarrow \infty}\int_{a_n}^{b} f(x)dx qed
Does anyone with more insight see how to do this more rigorously or is this the way?