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The dominated convergence theorem says :

Suppose $(f_n)$ is a sequence of measurable function s.t. $f_n(x)\longrightarrow f(x)$ a.e. as $n\to \infty $. If $|f_n(x)|\leq g(x)$ and $g$ is integrable, then $$\lim_{n\to\infty }\int f_n=\int f.$$

Quest 1) If $|f|$ is integrable, do we directly have $\lim_{n\to\infty }\int f_n=\int f$ or we need to have $|f_n(x)|\leq |f(x)|$ ?

Quest 2) If no, I have that $g$ is continuous with compact support (and thus integrable). Why the dominated convergence theorem allows me to conclude that $$\lim_{\delta\to 1}\int g(\delta x)dx=\int g(x)dx\ \ ?$$

Don't talks about change the variable $u=\delta x$ to prove it, it's not my question (I know how to prove this). My question is precisely : Why can we conclude using dominated convergence ? Because we don't necessarily have $|g(\delta x)|\leq g(x)$, and thus I'm a little annoyed by this argument.

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  • $\begingroup$ I add the symbol $\int $ in the Quest 1). I think it's what you wanted to say. Feel free to change it if I'm wrong. $\endgroup$ – idm Dec 29 '15 at 15:14
  • $\begingroup$ For Quest 1), yes, you need the condition $|f_n|\leq g$ for all $n$. Otherwise $\lim_{n\to\infty }\int f_n=\int f$ always which is not true. $\endgroup$ – idm Dec 29 '15 at 15:31
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For question 2:

If $g$ is continuous with compact support $K$ then $g$ is bounded; say $|g|\leq M$.

Since $K$ is compact, it is closed and bounded, and hence has finite measure.

Let's consider the functions $g(\delta x)$ for $\delta$ close to $1$. These functions are supported of $\frac{1}{\delta}K$, and $|g(\delta x)|\leq M$.

So what are these functions dominated by? Suppose

$$K\subset [-N,N],$$ for some large $N$. Then $g(\delta x)$ is dominated by the function

$$M\chi_{[-N^2,N^2]}$$which is integrable.

Finally since $g$ is continuous, $g(\delta x) \rightarrow g(x)$ as $\delta \rightarrow 1$.

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  • $\begingroup$ I don't think that your justification is totally correct. The proof of the fact that the $(x,\delta)\longmapsto g(\delta x)$ is supported on a compact doesn't look that obvious. In your proof $K_{1/\delta}$ is not independant of $\delta$, and I think it's a problem. $\endgroup$ – user301068 Dec 29 '15 at 16:28
  • $\begingroup$ @user301068 sure, but $\frac {1}{\delta}K$ is not too big. It is still compact, and you can contain within a large enough compact set that is independent of $\delta$. This is why I said "$\delta$ close to 1$". $\endgroup$ – GaussTheBauss Dec 29 '15 at 16:33
  • $\begingroup$ The bounding function needs to be chosen more carefully. If $K = [1,2]$, then $K$ and $1000K = [1000,2000]$ would be disjoint, and so $M\chi_{1000K}$ would not dominate $x \mapsto g(\delta x)$ for any $\delta \approx 1$. $\endgroup$ – epimorphic Dec 29 '15 at 20:56
  • $\begingroup$ @epimorphic. You are very correct. I think I know how to fix this. Let me give it another shot $\endgroup$ – GaussTheBauss Dec 29 '15 at 20:58
  • $\begingroup$ @epimorphic How about now? $\endgroup$ – GaussTheBauss Dec 29 '15 at 21:02

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