Is $\int_a^{b}f(x) dx = \lim_{k\rightarrow \infty } \int_a^{b_k}f(x)$? Let $f:[a,b] \rightarrow \mathbb{R}$ be integrable.
Let $b_k \subset [a,b]$ be a sequence such that $\lim_{k\rightarrow \infty} {b_k} = b$
Consider the integral $\int_a^{b}f(x) dx$.
Is it always the case that $$\int_a^{b}f(x) dx = \lim_{k\rightarrow \infty } \int_a^{b_k}f(x)?$$
If yes, why? If no, could you please provide a counterexample?
I feel that is the case, however, not sure how to prove it(or at least start my proof)
 A: I assume you are dealing with the Riemann integral. First recall that if $f$ is integrable on a closed interval $[a,b]$, then $f$ has to be bounded. Take $M=\sup_{x\in[a,b]} |f(x)|$,then
$$\lim_{k\to\infty}\left|\int_{b_k}^b f(x)dx\right|\leq\lim_{k\to\infty}\int_{b_k}^b Mdx=\lim_{k\to\infty}M(b-b_k)=0$$
Thus the result follows.
A: The above theorem is true. The integrability condition of the function and limit at posinfinity on sequence $b_k$ assures you that your both integrals are equal.
A: Using additivity property of definite integrals, if we can write  $\int_a^bf(x)dx=\int_a^{b_1}f(x)dx +\int_{b_1}^{b_2}f(x)dx+ \int_{b_2}^{b_3}f(x)dx +...+\int_{b_k}^{b_{k+1}}f(x)dx +... $$= a_1 + a_2 + a_3 + ... + a_{k+1}+... $  $=\sum_{i=1}^{\infty} a_i$. Now, the other expression $\lim_{k\rightarrow \infty } \int_a^{b_k}f(x)$ is nothing but the limit partial sums. But there is something I'm not sure of which is if the use of additivity property in the first step is valid  or not . Other than that this offers an intuitive explaination for why the statement must be true. Comments are welcome.
