I'm currently working on some practice exercises in preparation for an exam in a probability theory class and I am working through the dominated convergence section. I am now onto some problems which include a Dirac delta at a support point and I want to check if I'm approaching the problems correctly. From what I know/have been taught, when you have an integral with a Dirac delta at a support point, you simply solve the function at that support point. Thus, I've basically just combined the techniques of other dominated convergence problems I've been working through with this nuance. Any guidance on whether this is at all correct would be much appreciated.

1.) $\lim_{n\to\infty} \int_0^\infty \frac{r^n}{1+r^{n+2}} d\delta[1] (r)$

With $\delta[1]$ meaning the Dirac delta at support 1 (sorry, didn't know any better MathJax code for that).

Answer thus far:

Forgetting the Dirac delta for the moment, decompose the integral to $\int_0^1+ \int_1^{\infty}$, and for each integrand find a dominating function.

$$\int_0^{\infty}f_n(r)dr = \int_0^1 f_n(r)dr + \int_1^{\infty} f_n(r)dr$$

On $[0,1]$, $f_n(r)\to 0$ almost everywhere (it converges to $0$ except at $r=1$) and:

$$0\le f_n(r)\le \frac1{1+r^2} \in L^1$$

On $[1,\infty)$, $f_n(r)\to \frac1{r^2}$, and:

$$0\le f_n(r)\le \frac{1}{r^2} \in L^1$$

Thus, we have our dominated integrable functions and can use the theorem, i.e.

$\lim_{n\to\infty} \int_0^\infty \frac{r^n}{1+r^{n+2}} d\delta[1] (r)$ = $\int_0^\infty lim_{n\to\infty} \frac{r^n}{1+r^{n+2}} d\delta[1] (r)$

But again, considering the Dirac delta, don't we just solve the function at the support point. Thus, with r = 1 and n going to infinity, we get 0. Correct?

2.) $\lim_{n\to\infty} \int_{\mathbb R^2} e^{-(x^2 + y^2)^n} d\delta[C] (x,y)$

Where C is the unit circle centered at the origin.


Again, focusing on the first part of the integral, we can say that $\vert e^{-(x^2 + y^2)^n}\vert \le e^{-(x^2 + y^2)}$ integrable.

So, again, if we're just evaluating the function at the Dirac support, which is the unit circle, i.e. $x^2 + y^2 =1$, we get $1/e$.

  • $\begingroup$ ? If $\delta[1](r)$ means $\delta(r-1)$, then $\int_0^\infty f(r) \delta(r-1)dr = 1$ whenever $f$ is continuous at $r=1$, so that $\lim_{n \to \infty}\int_0^\infty \frac{r^n}{1+r^{n+2}} \delta(r-1)dr =1/2$ $\endgroup$ – reuns Nov 6 '16 at 22:38
  • $\begingroup$ Can you explain that? Again, I've been taught that an integral with respect to a dirac measure is simply the value of the function at the support point. $\endgroup$ – Archetupon Nov 7 '16 at 12:31
  • $\begingroup$ Yes, if the function is continuous, that's what I wrote. $\endgroup$ – reuns Nov 7 '16 at 20:53
  • $\begingroup$ And $\delta(r-1)$ is the (distributional, or Radon nikodym) derivative of the Dirac measure. So in every case, $d \delta(r)$ isn't an acceptable notation. The correct notation is $\int_{\mathbb{R}} f d \mu$ where $\mu$ is the Dirac measure at $1$ : $\mu(A) = 1$ if $1 \in A$, $\mu(A) = 0$ otherwise $\endgroup$ – reuns Nov 7 '16 at 20:57

These are pretty strange questions to test the dominated convergence theorem. In both problems, the integral does not actually depend on $n$.

  1. With the delta mass, you may as well just compute the integral directly and take the limit: $$\lim_{n\to\infty} \int_0^\infty \frac{r^n}{1+r^{n+2}}\,\delta_1 (dr) =\lim_{n\to\infty} \frac{1^n}{1+1^{n+2}}={1\over 2}.$$

  2. If the support of the measure is the unit circle, then you shouldn't call it a Dirac measure. Anyway, if the support of $\mu$ is on the unit circle then $$\lim_{n\to\infty} \int_{\mathbb R^2} e^{-(x^2 + y^2)^n} \mu(dx,dy)=\lim_{n\to\infty} e^{-1^n}=e^{-1}.$$

  • $\begingroup$ Great, thank you much. Also thanks to the poster above me, I get the 1/2 now, stupid oversight on my part. And in regards to them being "pretty strange"..haha welcome to my world right now! $\endgroup$ – Archetupon Nov 7 '16 at 16:33

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