General procedure to switch order of the integrals

Question

I am having problems in understanding how the limits of the integral change when we switch the order of integration. In particular, let $f,g : \mathbb{R} \rightarrow \mathbb{R}$ be integrable functions. How do I show that for $- \infty < a < b < + \infty$ the following is true \begin{align} \int_a^b \int_a^t f(s) g(t) ds dt = \int_a^b \int_s^b f(s) g(t) dt ds \end{align}

Answer 1

Your integration domain is $D := \lbrace (s,t) \in (a,b)^2: s < t\rbrace$.

Therefore, $$\int_a^b \int_a^t f(s)g(t) \,ds\,dt = \iint_D f(s)g(t) \,d(s,t) = \int_a^b \int_s^b f(s)g(t) \,dt\,ds.$$

If you let $t \in(a,b)$ arbitrary, you want $a < s < t$ to hold and if $s\in (a,b)$ is some real number, you want $s < t <b$.