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Prove that, given a sequence of measurable functions $\{f_{n}\}$, the set of points at which $\{f_{n}\}$ converge is measurable.

My solution is to first define $f(x) = \limsup_{n \to \infty} f_{n}(x)$, which is measurable. Then we also know all the differences $|f_{n}(x) - f(x)|$ are measurable, as well.

Now, I want to represent the set of points where these functions converge, but I am not entirely sure. I've reduced it to two possibilities, but am not sure which one is correct:

$$\bigcap_{k=1}^{\infty} \bigcup_{n=1}^{\infty} \{x : |f_{n}(x) - f(x)| < \frac{1}{k}\}$$

or

$$\bigcap_{k=1}^{\infty} \bigcup_{m=1}^{\infty} \bigcap_{n=1}^{\infty} \{x : |f_{n}(x) - f(x)| < \frac{1}{k}\}$$

mistertanaka
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  • I am a bit confused. What is exactly role of $m$ in second formula? It only appears as an index of the union, but $m$ itself is nowhere to be found inside the expression ${x: |f_{n}(x)-f(x)|<\frac{1}{k}}$. – Prism Apr 22 '14 at 22:51
  • By the way, see t.b.'s answer here. :) – Prism Apr 22 '14 at 22:54

1 Answers1

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We are looking to find the $x$ such that $f_n(x)$ is a Cauchy sequence, i.e.:

$$\{x \mid \forall \epsilon >0 :\exists n: \forall m \ge n: |f_n(x)-f_m(x)| < \epsilon\}$$

Translating the quantifiers to unions and intersections one by one, we obtain:

\begin{align*} & \{x \mid \forall \epsilon >0 :\exists n: \forall m \ge n: |f_n(x)-f_m(x)| < \epsilon\}\\ =& \bigcap_{k \ge 1} \left\{x \mid \exists n: \forall m \ge n: |f_n(x) - f_m(x)| < \frac 1k\right\} \\ =& \bigcap_{k\ge 1} \bigcup_{n \ge 1} \left\{x \mid \forall m \ge n: |f_n(x) - f_m(x)| < \frac1k\right\} \\ =& \bigcap_{k \ge 1} \bigcup_{n \ge 1} \bigcap_{m \ge n} \left\{x \mid |f_n(x) - f_m(x)| < \frac1k\right\} \end{align*}

In the formulation with the $\limsup$, this corresponds to your second option once you let the innermost intersection start at $n = m$.

To see that the first option is incorrect, consider any sequence which happens to satisfy $f_1 = f$.

Lord_Farin
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