# Hypothesis of dominated convergence theorem

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.

• 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. – idm Dec 29 '15 at 15:14
• 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. – idm Dec 29 '15 at 15:31

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 $\delta \rightarrow 1$.

• 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. – user301068 Dec 29 '15 at 16:28
• @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$". – GaussTheBauss Dec 29 '15 at 16:33 • The bounding function needs to be chosen more carefully. If$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\$. – epimorphic Dec 29 '15 at 20:56
• @epimorphic. You are very correct. I think I know how to fix this. Let me give it another shot – GaussTheBauss Dec 29 '15 at 20:58
• @epimorphic How about now? – GaussTheBauss Dec 29 '15 at 21:02

For question 1, to apply the dominated convergence theorem, you definitely need $$|f_{n}| \leq |f|$$ even if $$f$$ is integrable (as the first comment points out).

Here's a simple example to show why the integrability of $$f$$ is not enough. Let $$f_{n}(x) = 1$$ if $$x \in [n,n+1)$$, and $$0$$ otherwise. Here the domain of $$f_{n}$$ is $$\mathbb{R}$$. Then $$f$$ is identically equal to zero and obviously integrable, but $$1=\lim_{n} \int f_{n} > \int f = 0.$$

On the other hand, to show $$\lim_{n} \int f_{n} = \int \lim_{n} f_{n}$$, you don't have to use the dominated convergence theorem. There're some extensions of the theorem that do not require the existence of a dominating function. There's also a recent paper that gives a necessary and sufficient condition for $$\lim_{n} \int f_{n} = \int \lim_{n} f_{n}$$.

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