Let $\{f_n\}$ be a sequence of measurable functions on $X$, and suppose that, for every $x \in X$:

$$0 \le f_1(x) \le f_2(x) \le \cdots \le \infty $$ and $f_n(x) \rightarrow f(x) $ as $n \rightarrow \infty$ for every $x \in X$.

Let $s$ be any simple measurable function such that $0 \le s \le f$, let $c$ be a constant, $0 < c < 1$, and define

$$E_n = \{x : f_n(x) \ge cs(x) \} \ \ (n = 1,2,3 ...)$$

Then each $E_n$ is measurable. Why is this?

It would suffice to prove that the set $\{x : f_n(x) \ge cs(x) \}$ is open but this does not seem true to me.


1 Answer 1


Let $g_n = f_n - cs$. As $f_n$ and $s$ are measurable, so is $g_n$. Now note that

$$E_n = \{x \mid f_n(x) \geq cs(x)\} = \{x \mid f_n(x) - cs(x) \geq 0\} = \{x \mid g_n(x) \geq 0\} = g_n^{-1}([0, \infty)).$$

As $g_n$ is measurable and $[0, \infty)$ is a Borel set, $E_n = g_n^{-1}([0, \infty))$ is measurable.

  • $\begingroup$ We are in the extended reals setting so we would obtain $[0, \infty ]$ instead of $[0, \infty)$ would this be a problem? $[0, \infty]$ is no longer open? $\endgroup$
    – Monolite
    Dec 30, 2014 at 2:42
  • 1
    $\begingroup$ @Monolite: $[0, \infty)$ is closed in $\mathbb{R}$, not open. Regardless, we just need the set to be a Borel set. As $[0, \infty]$ is a Borel set in $\overline{\mathbb{R}}$, $g_n^{-1}([0, \infty])$ is measurable. $\endgroup$ Dec 30, 2014 at 2:49
  • $\begingroup$ Sorry to revive, but may you explain why $g_n = f_n-cs$ is measurable? I have given a proof here: math.stackexchange.com/questions/2070175/… but I am unsure. $\endgroup$
    – Bryan Shih
    Dec 26, 2016 at 3:11

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