4
$\begingroup$

Let $f_n:\mathbb{R}\rightarrow\mathbb{R}$ be $1/n$ for $x\in(0,n)$ and be $0$ otherwise. Then $f_n\rightarrow 0$ pointwise, and it is dominated by the function $g=1$, and $\int f_nd\mu=1$. So we have $\lim_{n\rightarrow\infty} \int f_nd\mu=1$ but $\int 0d\mu=0$. Why does this not violate the dominated convergence theorem?

$\endgroup$
5
$\begingroup$

I presume you mean the function that is $1/n$ on $(0,n)$ (otherwise this is not going to $0$ pointwise). The constant function $g=1$ is not integrable: it has integral $\infty$.

$\endgroup$
4
$\begingroup$

The functions $f_n$ are dominated by $g(x) = 1$, but $g$ is not integrable, i.e. $\int g \not< \infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.