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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.

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2 Answers 2

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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.

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    $\begingroup$ 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. $\endgroup$ Commented Nov 10, 2015 at 1:45
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The note suggests a slightly stronger definition of uniform integrability with the $3^{rd}$ criterion modified:

(No escape to width or horizontal infinity) For every ${\varepsilon>0}$, there is a finite measure subset ${E_\varepsilon}$ of ${X}$ such that ${\int_{X \backslash E_\varepsilon} |f_n|\ d\mu \leq \varepsilon}$ for all ${n}$.

In this case, with the dominated sequence of measurable function $f_n$, and the dominating function $g$, one can define $E_m := \{x \in X: g(x) \geq 1/m\}$ for $m = 1, 2, ...$ Then $E_m$ is increasing, $\mu(E_m) < \infty$ for each $m$ by Markov's inequality, and $X = \bigcup_{m=1}^\infty E_m$. By horizontal truncation, we then have: $\lim_{m \to \infty}\int_{E_m} g\ d\mu = \lim_{m \to \infty}\int_{X} g \cdot 1_{E_m}\ d\mu = \int_{X} g\ d\mu$.

In particular, $\forall \varepsilon > 0$, $\exists m \geq 1$ such that $\int_{X} g\ d\mu - \int_{E_m} g\ d\mu \leq \varepsilon$. Define $E_\varepsilon := E_m$. We see that for every $n$, $\int_{X \backslash E_\varepsilon} |f_n|\ d\mu \leq \int_{X \backslash E_\varepsilon} g\ d\mu \leq \varepsilon$, completing the proof.

The stronger criterion implies the original weaker one, by splitting the $L_1$ mass of $f_n$ in the set $|f_n| \geq M$ into $2$ parts: The part of mass concentrated in $E_\varepsilon$ and the part which isn't. Then control the 2 parts separately.

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