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.


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