Interchanging sum and integral in proof of summation by parts

Question

Why does this equality hold?

$$\sum_{y\leq n \leq x}a_n\int_n^x g'(t)dt = \int_y^x (\sum_{y\leq n\leq t}a_n)g'(t)dt$$

Answer 1

Note that your bounds are $y \leq n \leq x$ and $n \leq t \leq x$ where $x,y,t$ are (presumably) real numbers and $n$ an integer. Let's rewrite your sums so that the inner limit doesn't depend on the outer one. I use the notation $1_{condition}$ to be 1 if the condition is true, and 0 otherwise. $$ \sum_{y \leq n \leq x} a_n \int_n^x g'(t)\,dt = \sum_{y \leq n \leq x} a_n \int_y^x g'(t)1_{t \geq n} \,dt = \sum_{y \leq n \leq x} \int_y^x a_ng'(t)1_{t \geq n} \,dt. $$ Now notice that there are a fixed and finite number of $n$ between $y$ and $x$, so by linearity of the integral $\int(f+g) = \int f + \int g$ applied multiple times, we can always exchange a finite sum of Riemann-integrable functions (as your hypotheses imply) with an integral. $$ = \int_y^x \sum_{y \leq n \leq x}a_ng'(t) \,dt =\int_y^x \sum_{y \leq n \leq t}a_ng'(t)1_{t \geq n} \,dt. $$