Definition of measurable set: A set $E$ measurable if $$m^*(A) = m^*(A \bigcap E) + m^*(A \bigcap E^c)$$ for every subset of $A$ of $\mathbb R^n$.

Definition of Lebesgue measurable function: Given a function $f: D \to \mathbb R ∪ \{+\infty, -\infty\}$, defined on some domain $D \subset \mathbb{R}^n$, we say that $f$ is Lebesgue measurable if $D$ is measurable and if, for each $a\in[-\infty, +\infty]$, the set $\{x\in D \mid f(x) > a\}$ is measurable.

Definition of almost everywhere convergence: Suppose $f(x), f_1(x), f_2(x), \dots, f_k(x), \dots$ are extended real functions defined on a set $E \subset \mathbb R^n$ that is each $f_i: E \to [-\infty, +\infty] $. If $\exists Z \subset E$ such that $m(Z)=0$ and $$\lim_{k \to +\infty}f_k(x) = f(x), x\in(E-Z),$$ then $\{f_k(x)\}$ converges almost everywhere to $f(x)$ on $E$, namely $$f_k(x) \to f(x),\ \text{a.e.}\ x ∈ E.$$

I think I can translate "$f_k(x) \to f(x),\ \text{a.e.}\ x \in E$" into an exact math symbol that should be $m(\{x\in E \mid(\lim_{k\to+\infty}f_k(x)) \neq f(x)\}) = 0$. However, I'm not sure whether the "$\lim$" can be taken out of the set that is

$$m\left(\lim\limits_{k\to+\infty}\{x\in E \mid f_k(x) \neq f(x)\}\right)=0?$$

I have two hurdles ahead, one is $\left\{x\in E \mid \lim\limits_{k\to+\infty}f_k(x) \neq f(x)\right\}$ may not be equal to $\lim\limits_{k\to+\infty} \{x\in E \mid f_k(x) \neq f(x)\}$ and the other is $\lim\limits_{k\to+\infty} \{x\in E \mid f_k(x) \neq f(x)\}$ may not exist.

I tried an example on the domain $[0, 1]$: $f_k(x) = x^k$ and $f(x) = 0$. I have $f_k(x) \to f(x),\ \text{a.e.}\ x \in [0,1]$ is equivalent to

\begin{align*} \left\{x\in[0,1] \mid \lim\limits_{k\to+\infty}f_k(x) \neq f(x)\right\} &= \{1\}\\ &\neq \lim\limits_{k\to+\infty} \{x\in [0,1] \mid f_k(x) \neq f(x)\}\\ &= \lim\limits_{k\to+\infty}(0,1]\\ &= (0,1] \end{align*}

and $m(\{1\}) \neq m((0,1])$. Even,

\begin{align*} m\left(\left\{x\in[0,1] \mid \lim\limits_{k\to+\infty}f_k(x) \neq f(x)\right\}\right) &= m(\{1\})\\ &= 0\\ &\neq \lim_{k\to+\infty} m(\{x\in [0,1] \mid f_k(x) \neq f(x)\})\\ &= \lim_{k\to+\infty}m((0,1])\\ &= 1\\ &= m\left(\lim\limits_{k\to+\infty} \{x\in [0,1] \mid f_k(x) \neq f(x)\}\right). \end{align*}

So from this example, I know

$$m\left(\left\{x\in [0,1] \mid \lim\limits_{k\to+\infty}f_k(x) \neq f(x)\right\}\right) \neq \lim\limits_{k\to+\infty} m(\{x\in[0,1] \mid f_k(x) \neq f(x)\})$$


$$m\left(\left\{x\in [0,1] \mid \lim\limits_{k\to+\infty}f_k(x) \neq f(x)\right\}\right) \neq m\left(\lim\limits_{k\to+\infty} \{x\in [0,1] \mid f_k(x) \neq f(x)\}\right).$$

But if I replace $\lim\limits_{k\to+\infty} \{x\in [0,1] \mid f_k(x) \neq f(x)\}$ with

$$E(n,k) = \bigcup_{m=n}^{\infty}\left\{x\in [0,1] : |f_m(x) - f(x)| \geq \frac{1}{k}\right\}$$

where $k\in\mathbb N^+$, then

$$m(E(n,k)) = m\left(\bigcup_{m=n}^{\infty}\left[\left(\frac{1}{k}\right)^{\frac{1}{m}}, 1\right]\right) =m\left(\left[\left(\frac{1}{k}\right)^{\frac{1}{n}}, 1\right]\right) \to 0$$

as $n\to+\infty$ for each $k$.

What happened here? Why does exchanging order of $\lim$ and measure will cause such a big difference? And why can union operation keep the measure the set have? Or is there something wrong with my idea or example?

  • $\begingroup$ Update: It is not appropriate to write out $\lim_{k\to+\infty} m(\{x\in [0,1] \mid f_k(x) \neq f(x)\})$ coz its definition is totally different from definition of $\lim_{k->+\infty}f_k(x)$. That's my mistake. $\endgroup$ – Bear and bunny Jul 9 '15 at 2:58

$\{x:\lim f_k(x)\ne f(x)\}$ simply has nothing to do with the limit of the sets $\{x:f_k(x)\ne f(x)\}$. For example let $f_k(x)=1/k$ for all $k$. Why doesn't this have anything to do with that? Well, it doesn't - you don't have any good reason to think it does.

  • $\begingroup$ seems i got your idea. Definition of $lim_{k->+\infty}\{x:f_k(x)\ne f(x)\}$ is totally different from the definition we used for $lim_{k->+\infty}f_k(x)$. $\endgroup$ – Bear and bunny Jul 8 '15 at 18:48
  • 1
    $\begingroup$ No. But if you change the union to an intersection then yes... $\endgroup$ – David C. Ullrich Jul 8 '15 at 19:23
  • 1
    $\begingroup$ Almost. A limit can equal $1/k$ even though all the terms are strictly less than $1/k$. But if a limit if greater or equal to $1/k$ then there must be infinitely many terms larger than, say $1/(2k)$. So this would be correct if you changed the very last $1/k$ to $1/(2k)$ (or anything else depending only on $k$ and smaller than $1/k$). $\endgroup$ – David C. Ullrich Jul 8 '15 at 19:38
  • 1
    $\begingroup$ You asked whether a certain formula was true. There's that long display. I'm not going to retype it all. In that display I see $\frac1k$ in three places. There's the first place I see, then the second place, then the third. If you change the third one to $\frac1{2k}$ it becomes correct $\endgroup$ – David C. Ullrich Jul 8 '15 at 19:46
  • 1
    $\begingroup$ If "when $k>N$" means "for all $k>N$ then yes. $\endgroup$ – David C. Ullrich Jul 9 '15 at 16:06

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.