Question on additivity of Riemann Integral . Given a Riemann integrable function , $f:[a,b] \rightarrow \mathbb{R}$ and a sequence ${b_k\to b}$  ( $b_k \in [a,b]$ ) is the following use of additivity of integrals correct ?  $\int_a^bf(x)dx=\int_a^{b_1}f(x)dx +\int_{b_1}^{b_2}f(x)dx+ \int_{b_2}^{b_3}f(x)dx +...+\int_{b_k}^{b_{k+1}}f(x)dx +...$   I realized this issue when I was giving an answer here : Is $\int_a^{b}f(x) dx = \lim_{k\rightarrow \infty } \int_a^{b_k}f(x)$?  Also, what happens if we talk about the same issue in terms of lebesgue integration ?

The definition of the Riemann integral ensures that if $f$ is Riemann integrable on the interval $[a,b]$ then it is also Riemann integrable on any subinterval. It also ensures that $f$ is bounded, so assume $|f(x)|\leq M$ for all $x\in[a,b]$. Then, for each $k$ we have
$$\int_a^b f(x)\ dx-\int_{b_k}^bf(x)\ dx = \int_a^{b_1}f(x)\ dx+\int_{b_1}^{b_2}f(x)\ dx+\cdots+\int_{b_{k-1}}^{b_k}f(x)\ dx$$
Now, we have $|\int_{b_k}^bf(x)\ dx| \leq (b-b_k)M \to 0$. Thus the sum on the right converges to $\int_a^b f(x)$, as desired.
The same result is true if $f$ is Lebesgue integrable:  set $f_k=\chi_{[a,b_k]}f$. Then
$$\int_a^b f_k(x)\ dx = \int_a^{b_k} f(x)\ dx =  \int_a^{b_1}f(x)\ dx+\int_{b_1}^{b_2}f(x)\ dx+\cdots+\int_{b_{k-1}}^{b_k}f(x)\ dx$$
The left side converges to $\int_a^b f(x)\ dx$ by the dominated convergence theorem, which proves the desired result.
