Good day, under the definitions:

Let the sequence of measurable functions $f_{i}(x)$ be defined and finite almost everywhere on a measurable set $E$. Let $g(x)$ be a measurable function which is finite almost everywhere. If $\lim_{n\rightarrow \infty}m(E\cap (|f_n-g|\geq t))=0$, for all positive numbers $t$, then the sequence is said to converge in measure to the function $g(x)$.

Let $M = \left \{ f(x) \right \}$ be a family of Lebesgue integrable functions defined on a set $E$. If for every $\varepsilon > 0$ there exists a $\delta > 0$ such that the relations $e\subset E$, $me< \delta$ imply $|\int_{e}f(x)dx|< \varepsilon $ for all functions of the family $M$, then the functions of the family $M$ are said to have equiabsolutely continuous integrals.

We have the Vitali theorem: Let a sequence of Lebesgue integrable functions ${f_i(x)}$ converging in measure to $g(x)$, be defined on a measurable set $E$. If the functions of the sequence have equiabsolutely continuous integrals, then $g(x)$ is Lebesgue integrable function and $\lim_{n\rightarrow \infty} \int_{E}f_n(x)dx=\int_{E}g(x)dx$.

I need an example such that if the functions of the sequence have not equiabsolutely continuous integrals, the theorem of Vitali does not apply.

I I doubt if I can work it with the variants $\int_{e}|f_n(x)|dx, \int_{e}|g(x)|dx$?

  • 2
    $\begingroup$ Consider $f_n = n\chi_{[0,1/n]}$ on $[0,1].$ $\endgroup$
    – zhw.
    Sep 16, 2017 at 15:55

1 Answer 1


This version of the Vitali theorem is only true if your set $E$ has finite measure. Otherwise take $E = \mathbb{R}$, $f_n = 1_{[n, n+1]}$, $g=0$. These functions have equiabsolutely continuous integrals (take $\delta = \epsilon$) but the conclusion fails.

Anyway, assuming $E$ has finite measure, just take your favorite example of a sequence that converges in measure but not in $L^1$. Every measure theory student has got to know such an example. If you don't already, then try $E = [0,1]$, $f_n = n 1_{[0,1/n]}$ as zhw suggested in a comment. We have $\int f_n = 1$ but $f_n \to 0$ in measure, so the conclusion of Vitali fails, and if you believe it is a theorem then the hypothesis has to fail, i.e. this sequence does not have equiabsolutely continuous integrals. But it is also a good exercise to verify that directly.


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.