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 $a,b\in\mathbb R$, $a<b$,

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

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 $|F_n(x)|<h$, $$ |F_n(x)| \le \int_{a_n}^{b_n} |f(x,t)| dt \le \int_{-\infty}^{\infty} |f(x,t)| dt \le h(x)\in L^1.$$ 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.

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

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