1
$\begingroup$

I want to find some sequence $\{f_n\}$ of Lebesgue integrable functions on an interval $I$ such that the pointwise limit exists almost everywhere on $I$, the limit function $f$ is Lebesgue-integrable on $I$ but: $$\int_I f \neq \lim_{n\to\infty}\int_I f_n.$$ In Dominated Convergence Theorem we have that if $f_n$ is dominated by some Lebesgue integrable function $g$ we don't even need the hypothesis of $f$ being Lebesgue-Integrable. But what happens if we add it but remove the hypothesis of $g$ dominating the sequence?

$\endgroup$
1
  • 1
    $\begingroup$ Let $f_n=n1_{[0,1/n]}$, then $f_n$ conveges to $f=0$, but $\int f_nd\mu=1$. $\endgroup$ Commented Apr 18, 2017 at 0:51

3 Answers 3

3
$\begingroup$

Let the underlying space be $(0,1)$ with Lebesgue measure. Let $f_n=n\cdot 1_{(0,1/n)}$. Then $f_n\to 0$ pointwise, but $\int_0^1 f_n=1$.

$\endgroup$
2
  • $\begingroup$ You mean pointwise a.e. $\endgroup$
    – zhw.
    Commented Apr 18, 2017 at 1:07
  • $\begingroup$ @zhw. Corrected, thanks. $\endgroup$ Commented Apr 18, 2017 at 1:09
2
$\begingroup$

You can always look at the classical $f_n(x) = nx^n$ on $(0,1).$

$\endgroup$
2
$\begingroup$

For an example consisting of continuous functions on a closed interval and pointwise convergence everywhere (to a continuous function), consider $f_n:[0,1]\to \mathbb{R}$ given by a graph of a triangle with one foot on $0$, basis $\frac{1}{2^n}$ and height $2^n$, and $0$ everywhere else.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .