Let $f:[a,b] \to \mathbb{R}$ be an Riemann integrable function and let ${b_n}$ be a sequence in $[a,b]$ such that $$\lim_{n\to\infty}b_n=b.$$ Show that $$\lim_{n\to\infty}\int_a^{b_n}f(x)dx=\int_a^bf(x)dx.$$
Attempt at solution:$$\begin{equation}\begin{split}\lim_{n\to\infty}\int_a^{b_n}f(x)dx & =\lim_{\|P\|\to0}\lim_{n\to\infty}\sum_{k=1}^{m}f(\xi_k)\Delta x_k\\ & =\lim_{\|P\|\to0}\left (\sum_{k=1}^{m-1}f(\xi_k)\Delta x_k +\lim_{n\to\infty}f(\xi_m)(b_n-x_{m-1}) \right ) \\ & = \lim_{\|P\|\to0}\left (\sum_{k=1}^{m-1}f(\xi_k)\Delta x_k +f(\xi_m)(b-x_{m-1}) \right ) \\ & = \int_a^bf(x)dx. \end{split}\end{equation}$$ It seems a bit simple. Is there something I'm missing?
Hint: $\int_a^b f - \int_a^{b_n} f = \int_{b_n}^b f .$