# Integration of function of two variables

I'm studying Measure theory and Integration and i found this problem at Bartle's book "Elements of Integration". I understand that we have to use the Dominated Convergence Theorem but i don't see the way. Any advice\hint it would be very helpful.

Suppose the function $$x\to f(x,t)$$ is $$X$$-measurable for each real number $$t$$, and the function $$t\to f(x,t)$$ is continuous on R for each $$x\in X$$. In addition suppose that there integrable functions $$g, h$$ on $$X$$ such that $$|f(x,t)|\le g(x)$$ and such that the improper Riemann integral satisfies the inequality $$\int_{-\infty}^{\infty} |f(x,t)| \mathrm{d}t \le h(x).$$ Show that $$\int_{-\infty}^{\infty} \left[\int_X f(x,t) \mathrm{d}\mu(x) \right] \mathrm{d}t= \int_X \left[\int_{-\infty}^{\infty} f(x,t) \mathrm{d}t \right] \mathrm{d}\mu(x),$$ where the integrals with respect to t are improper Riemann integrals.

We first show that for any $$a,b\in\mathbb R$$, $$a,

$$\int_{a}^{b} \int_X f(x,t) dμ(x) dt= \int_X \int_{a}^{b} f(x,t) dt dμ(x)$$

For LHS to make sense:

• $$(A)f(x,t)$$ needs to be measurable and integrable for every $$t$$
• $$(B)$$ $$\int_X f(x,t)d \mu(x)$$ needs to be Riemann integrable on $$[a,b]$$

$$(A)$$ is covered by the measurability assumption and the existence of $$g$$. For $$(B)$$, we have continuity in the integral by dominated convergence- if $$t_n\to t$$, then $$f(x,t_n)\to f(x,t)$$ pointwise in $$x$$ (using continuity in $$t$$), and $$\sup_n |f(x,t_n)| < g(x)$$, $$g\in L^1$$, so $$\int_X f(x,t_n) d\mu \to \int_X f(x,t) d\mu .$$ A continuous function on $$\mathbb R$$ is Riemann integrable on every $$[a,b]$$.

For RHS to make sense:

• $$(C)$$ $$f(x,t)$$ needs to be Riemann integrable on $$[a,b]$$ for almost every $$x$$
• $$(D)$$ $$\int_{a}^b f(x,t)dt$$ needs to be measurable and integrable

$$(C)$$ is covered by the continuity assumption. For $$(D)$$, its measurable because for each $$x$$, we can write it as the limit of the following functions, which are Riemann sums in $$t$$ (from If f is Riemann integrable on [a,b] then prove that there is an equispaced partition Pn such that limU(Pn,f) = limL(Pn,f) = the value of the integral. , we can choose the uniform grid mesh)

$$R_n[f](x) := \sum_{i=1}^n f\left(x,a+\frac{i(b-a)}n \right) \frac{b-a}n\to \int_{a}^b f(x,t)dt$$ Integrability is by the bound $$\int_X\left|\int_{a}^b f(x,t)dt\right| d\mu(x) \le \int_X\int_{a}^b |f(x,t)|dt d\mu(x)\le \int_X\int_{\mathbb R} |f(x,t)|dt d\mu(x) \le \int h d\mu < \infty$$

For the equality - Note that $$R_n \left[ \int_X f d\mu\right ] = \int_X R_n f(x) d\mu(x)$$ By the Riemann integrability of $$\int_X f d\mu$$, the LHS converges to $$\int_{a}^{b} \int_X f(x,t) dμ(x) dt$$. For the RHS, note that from $$\sup_{t} |f(x,t)| \le g(x)$$,

$$|R_n f(x)| \le \sum_{i=1}^n \left| f\left(x,a+\frac{i(b-a)}n \right)\right| \frac{b-a}n \le (b-a) g(x) \in L^1(X)$$ thus by Dominated convergence, the RHS converges to $$\int_X \int_{a}^{b} f(x,t) dt dμ(x)$$.

To finish, we now use the assumption on the improper integral. The convergence of the improper integral

$$F(x) = \lim_{\substack{a\to-\infty\\ b\to+\infty}} \int_a^b f(x,t) dt$$

is equivalent to the assertion that for any two sequences $$a_n\to-\infty$$ and $$b_n\to\infty$$, $$F_n(x) := \int_{a_n}^{b_n} f(x,t) dt \to F(x).$$ The assumption $$\int_{\mathbb R} |f(x,t)| dt < h(x)$$ implies that $$|F_n(x)|, $$|F_n(x)| \le \int_{a_n}^{b_n} |f(x,t)| dt \le \int_{-\infty}^{\infty} |f(x,t)| dt \le h(x)\in L^1.$$ Thus, by the equality for each fixed $$[a,b]$$ just proved and then dominated convergence,

\begin{align} \lim_{n\to\infty}\int_{a_n}^{b_n} \int_X f(x,t) dμ(x) dt &= \lim_{n\to\infty}\int_X \int_{a_n}^{b_n} f(x,t) dt dμ(x) \\ &= \lim_{n\to\infty} \int_X F_n(x) dx \\ &= \int_X F(x) dx \\ &= \int_X \int_{\mathbb R} f(x,t) dμ(x) dt \end{align} Thus, the improper integral $$\int_{\mathbb R} \int_X f(x,t) dμ(x) dt$$ exists, and equals the claimed value.

• Hope its clear, if you see something odd, do tell me – Calvin Khor Sep 6 '19 at 11:18