I think I am getting a little better at these MCT, DCT-type exercises. The issue is to show/prove the existence and finiteness (if they apply) to the following function:

$$f(x)=\sin \left(\dfrac1{x^2} \right)$$

Where applicable I want to show as rigorously as possible the justification for existence, etc. For example, it is not enough to say simply, "this exists by MCT"; rather, I need to show that the conditions of measurability, monotone-increasing, etc. are met.

Work/attempt at a solution:

First, although I have a limited experience with this type of integral, I believe the function can be re-written as follows:


If I have this correctly (I attempted to extrapolate from an integral table I found regarding $\sin\left(\dfrac1x \right)$, but please correct me if I am wrong), I can utilize Dominated Convergence to show the integral exists. That is,

If $f_1,f_2,...$ are measurable functions and $|f_n|\le g$ for $\mu$-integrable function, $g$, and if $f_n\to f$, $\mu$-a.e, then $\lim_{n\to\infty}\int f_n d\mu\to\int fd\mu$.

First, I need to show measurability. I believe that continuity on the interval in question, $(0,\infty)$, is sufficient to show measurability, correct?

Next, define

$$S_n(x):=\sum_{n=0}^{t}\frac{(-1)^n}{x^{2(2n+1)}(2n+1)!}\to f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{x^{2(2n+1)}(2n+1)!}$$

Then, define

$g(x)=\dfrac1x\ge |f_n|,\forall x>0$

and therefore conclude by Dominated Convergence Theorem, that $\displaystyle \int_{(0,\infty)}\sin \left(\dfrac1{x^2} \right)d\mu$ exists.

Please comment, add answers, critique, tell me where I made mistakes, etc. I am learning this on my own so any and all improvements are welcome.

From here I need to show whether this integral is finite or infinite. Is there a way to show this without an explicit calculation of the integral?


1 Answer 1


$\sin(1/x^2)$ is continuous in $(0,\infty)$ and hence it is also measurable.

Split your domain $(0,\infty)$ into $(0,1] \cup (1,\infty)$.

In the interval $(0,1]$, bound $\sin(1/x^2)$ by $1$ and in the interval $(1,\infty)$, bound $\sin(1/x^2)$ by $1/x^2$ and then argue out.

  • $\begingroup$ why bother to split the interval as $\sin(\frac{1}{x^2})$ $\leq 1$ $\forall x \in (0,\infty)$? $\endgroup$
    – miyagi_do
    Dec 26, 2017 at 15:52

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