5
$\begingroup$

This is an exercise from tao's blog.

A sequence $f_n:X\to\mathbf{C}$ of absolutely integrable functions is said to be uniformly integrate if the following three statements hold:

  1. (Uniform bound on $L^1$ norm) One has $\sup_n\|f_n\|_{L^1(\mu)}=\sup_n\int_X |f_n|\,d\mu<\infty$.
  2. (No escape to vertical infinity) One has $\sup_n\int_{|f_n|\geq M}|f_n|\,d\mu\to 0$ as $M\to \infty$.
  3. (No escape with width infinity) One has $\sup_n\int_{|f_n|\leq\delta}|f_n|\,d\mu\to 0$ as $\delta\to 0$.

Show that every dominated sequence of measurable functions is uniformly integrable.

Let $(f_n)_{n=1}^\infty$ is a sequence of measurable function dominated by some absolutely integrable function $g$, that is $|f_n|\leq g$ for all $n=1,2,\dots$. The first statement is trivial. Note that $$\int_{|f_n|\geq M}|f_n|\,d\mu\leq \int_{|f_n|\geq M}g\,d\mu\leq\int_{|g|\geq M}g\,d\mu,$$ then using dominated convergence theorem, the second statement follows. Now I have trouble in verifying the third statement, since $g\leq\delta$ is contained in $|f_n|\leq\delta$, the above argument doesn't work.

$\endgroup$
2
$\begingroup$

For arbitrary $m \in \Bbb{N}$, let $$ g_m := \sup_{n \in \Bbb{N}} (1_{|f_n| \leq 1/m} \cdot |f_n|). $$ Note that this is a measurable function with $0 \leq g_m \leq 1/m$, so that $g_m \to 0$ almost everywhere as $m \to \infty$.

Furthermore, it is easy to see $0 \leq g_m \leq \sup_n |f_n| \leq g$.

Thus, the dominated convergence theorem implies $$ \sup_n \int_{|f_n|\leq 1/m}|f_n| \, d\mu \leq \int \sup_n 1_{|f_n|\leq 1/m} \cdot |f_n| \, d\mu = \int g_m \, d\mu \xrightarrow[m\to\infty]{} 0. $$

By monotonicity (or since we can replace $(1_m)_m$ by any null-sequence), this implies the claim.

$\endgroup$
  • 1
    $\begingroup$ Thanks for you answer, for an alternative solution, we may control $|f|_n$ by $\min(g,\delta)$, that is $$\int_{|f_n|\leq\delta}|f_n|\,d\mu\leq\int_{|f_n|\leq\delta}\min(g,\delta) \,d\mu \leq\int_X\min(g,\delta)\,d\mu.$$ Note that $\min(g,\delta)$ goes zero as $\delta\to 0$ and dominated by $g$, then using dominated convergence theorem. $\endgroup$ – Xiang Yu Nov 10 '15 at 1:45

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.