If $m(E) < \infty,$ ${f_n}$ is uniformly bounded in $E,$ and $f_n(x) \to f(x)$ in $E,$ then $$ \lim_{n \to \infty} \int_E f_n\,dx = \int_E f\,dx$$

The statement is a corolary for which theorem?

I really don't know for sure, but the monotone convergence theorem states something close to that, but requiring that the $f_n's$ are monotonic. So maybe if the $f_n's$ are uniformly bounded and convergent, could that imply be that the $f_n's$ are monotonic (increasing or dereasing)?

Or maybe I could use a hint to how I could prove this?


  • $\begingroup$ You should say $f_n\to f$ pointwise on $E.$ $\endgroup$
    – zhw.
    Oct 28, 2017 at 5:21

1 Answer 1


The one-line proof of this is "Dominated Convergence Theorem." You can prove it in other ways:

By Egorov's theorem, pick a measurable set $A_\epsilon\subset E$ with the properties that $m(E-A_\epsilon) < \epsilon$ and $f_n$ converges to $f$ uniformly on $A_\epsilon$. Then write $$ \int_E |f_n-f| = \int_{A_\epsilon}|f_n-f| + \int_{E-A_\epsilon}|f_n-f|, $$ and inspect each integral on the right-hand side as $n$ gets large.

If you don't have access to Egorov's theorem, then let $\epsilon>0$ and put $E_{k} = \{x \in E : |f_n(x)-f(x)| < \epsilon\ \text{for all $n\ge k$}\}$. The sets $E_k$ are increasing: $E_1\subset E_2\subset\dots$, and since $f_n\to f$ pointwise, their union is equal to $E$: $\bigcup_{k=1}^\infty E_k = E$. Hence $m(E_{k})\to m(E)$ as $k\to\infty$.

Now write $$ \int_E |f_n - f| = \int_{E_{k}}|f_n-f| + \int_{E-E_k}|f_n-f| $$ and look at what happens to these latter two integrals when $k$ and $n$ get large.


You must log in to answer this question.

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