# The applicability of the Dominated Convergence theorem on the real line

Let $f_n(x)=\frac{1}{n}\chi_{[0,n]}(x)$, $x\in\mathbb{R}$, $n\in\mathbb{N}$ and $\chi$ is the characteristic/indicator function. Now it is clear that $f_n\rightarrow 0$, but in the text I am using it says we can't apply the Dominated Convergence theorem as there is no function to dominate $f_n$

I have trouble seeing that. To me the $f_n$'s are dominated by the constant function $1$. So I think can apply the Dominated Convergence theorem in some cases which depends on the choice measure where the constant functions are integrable. So if we are using the Lebesgue measure (on $\mathbb{R}$) then we can't use the Dominated Convergence theorem (and the text is right), but if are using a finite measure then will it be valid to use the Dominated Convergence theorem and we will then have $\lim\limits_n\int f_n\rightarrow0$. Also if our space was just a bounded interval and we have the Lebesgue measure (since it is now a finite measure on the internal), then will it be fine to use the Dominated Convergence theorem on the $f_n$'s ? Is this right or wrong?

• $1$ is not integrable on $\mathbb{R}$ Jul 4, 2014 at 20:02
• Yes, I know that constant functions are not integrable on $(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu)$ if $\mu$ is the Lebesgue measure, but the constant functions integrable if $\mu$ was a probability or any finite measure on $\mathbb{R}$. My question was about when we can apply the Dominated Theorem. Jul 4, 2014 at 22:57

The smallest thing that could possibly dominate the functions $f_n$, $n=1,2,3,\ldots$ is the function $$x\mapsto g(x) = \sup_n |f_n(x)| = \begin{cases} \dfrac{1}{\lceil x \rceil} & \text{if }x>0, \\[10pt] 0 & \text{if }x<0. \end{cases}$$ The problem is that $$\int_{[0,\infty)} g(x)\,dx = \sum_{n=1}^\infty \frac 1 n = \infty.$$
Yes, if we are working only on the space $[0,1]$, for example, then $f_n$ restricted to this interval is dominated by $1$, which is integrable on $[0,1]$. So the DCT applies, and tells us that the limit of the integral is the integral of the limit, which is $0$. This is easily verified, since $f_n$ integrates to $1/n$ in this case, and $1/n \to 0$.
But as you said, if we are working on an unbounded interval and using Lebesgue measure, then $1$ is NOT integrable, and cannot be used as a dominating function.