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I'm trying to solve the following problem from the problem set of a Real Analysis course:

Let $E \subseteq \mathbb R^m$ be a measurable set (w.r.t the Lebesgue measure $\mu$) and $(f_k)_{k \in \mathbb N}$ a sequence of measurable functions from $E$ to $\mathbb R$ that are pointwise bounded. That is, $\displaystyle \forall x \in E \; \exists M_x >0 \; : \; \sup_{k \in \mathbb N} |f_k(x)| \le M_x$

Prove that if the sequence satisfies the following property:

$$\forall a > 0 \; \exists k_a \in \mathbb N \; : \; \forall k \ge k_a \; \mu \{|f_k| < a\} \le \frac a k$$

then $\mu(E) = 0$

What I've tried

We have for an arbitrary $a > 0$ and for all $k$ that $E = \{|f_k| < a\} \cup \{|f_k| \ge a\}$, which then implies $\mu(E) = \mu\{|f_k| < a\} + \mu\{|f_k| \ge a\}$

The stated property tells us that $\mu\{|f_k| < a\} \xrightarrow{k \to \infty} 0$ so we get that $\mu\{|f_k| \ge a\} \xrightarrow{k \to \infty} \mu(E)$

So one way of proving this would be to prove that such a limit converges to 0

In my head a possibility here is, for each k, using Lusin's theorem to get a closed set that approximates E arbitrarily in measure and f_k is continuous on that set. If such a set was compact then I can get a maximum for f_k on this set, but this isn't true in general.

Is there anything else I could try here?

Edit: so perhaps there is a subset of $E$ that arbitrarily approximates E in measure such that the sequence of functions is uniformly bounded on that set? That would work but I have no idea if it's true

Francisco José Letterio
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1 Answers1

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The assumption you haven't used is that $\{f_k\}_k$ is pointwise bounded.

Let's apply it.

$$E=\bigcup_{Q\in\mathbb{Q}^+}{\{x:M_x\leq Q\}}=\bigcup_{Q\in\mathbb{Q}^+}{\bigcap_k{\{x:f_k(x)<Q\}}}$$

Each term in the union is now controlled by the Markov-type hypothesis. To be precise, for cofinitely many $K$, $$\mu\left(\bigcap_k{\{x:f_k(x)<Q\}}\right)\leq\mu(\{x:f_K(x)<Q\})\leq\frac{Q}{K}$$ Sending $K\to\infty$, the intersection $$\bigcap_k{\{x:f_k(x)<Q\}}$$ is null.

But $E$ is a countable union of such intersections, and so itself null.

Jacob Manaker
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