I came across this (seemingly?) weird behavior of a sequence of simple functions:

Let $E$ be the Smith–Volterra–Cantor set and $m: \operatorname{Leb}(\Bbb{R}) \to [0, \infty]$ the Lebesgue-measure. Now, because $m(E) > 0$, there exists a non-measurable subset $F$ of $E$.

Using the notation for the construction process of the Cantor set used here, we see that if $x \in E$, then $$ \{x\} = \bigcap_{j=1}^\infty I_{j,k(x,j)} $$ for some sequence $(k(x,j))_{j \in \Bbb{N}}$ depending on $x$.

Now, we can write the set $F$ as $$ F = \bigcup_{x \in F} \bigcap_{j=1}^\infty I_{j,k(x,j)}\,. $$ Define a sequence of sets $(F_m)_{m \in \Bbb{N}}$ by $$ F_m = \bigcup_{x \in F} \bigcap_{j=1}^m I_{j,k(x,j)}\,. $$ and a sequence of functions $(f_m)_{m \in \Bbb{N}}$ by $f_m(x) = \chi_{F_m}(x)\,,$ where $\chi_{F_m}$ is the characteristic function of $F_m$.

If $m$ is fixed, there are only a finite amount of intervals $I_{j,k(x,j)}$, $1 \leq j \leq m$. This means that $F_m$ is a finite union of finite intersections of closed intervals, hence closed. So the functions $f_m$ are measurable.

Now, my intuition says that $f_m \to \chi_F$ pointwise, but this is impossible since $\chi_F$ is not measurable function.

My question is why $f_m \not \to \chi_F$? Is there some insight into this? Or are non-measurable sets just weird? Do I have a mistake somewhere?

Some notes: If $x \in F$, then $x \in F_m$ for all $m$, hence $f_m|_F \to \chi_F|_F$. So $f_m|_{F^c} \not \to \chi_F|_{F^c}$.

Because $\chi_F$ is $0$ on $F^c$, there must exist a point $ x \in F^c$ s.t. $f_m(x) \not \to 0$. So for all $n_0 \geq 1$ there exists $n \geq n_0$ s.t. $f_n(x) = 1$ which just means that $x \in F_n$. So no matter how far in the sequence we go, $x$ belongs to some $F_n$. Because $F_{n+1} \subset F_n$, this means that $x$ belongs to $F_n$ for all $n$?

The set $F$ can't be closed, so it has a limit point in $F^c$...

  • $\begingroup$ I am somewhat hesitant to accept your initial formula; a little more caution might be good there. That said, I think you are on the right track. In particular I think you will find that $\overline{F} \cap F^c = \partial F \cap F^c$ has positive outer measure. $\endgroup$
    – Ian
    Apr 6, 2015 at 12:32

1 Answer 1


Since $F$ can't be closed, it has a limit point in $F^c$.

Because $E$ is closed and $F \subset E$, the limit point of $F$ must be in $E \cap F^c$. Now let $x$ be this limit point. Since $x \in E$, we can write $$ \{x\} = \bigcap_{j=1}^\infty I_{j, k(x,j)} $$ where $(k(x,j))_{j \in \mathbb{N}}$ is some sequence. Since $x$ is a limit point of $F$, for all $n$ there exists $y_n \in F$ such that $$ \{y_n\} = \bigcap_{j=1}^\infty I_{j, k(y_n,j)} $$ and $|x-y_n| < 1/n$. This then implies that \begin{align} x \in \bigcap_{j=1}^{r_n} I_{j, k(y_n,j)} \end{align} for some $r_n$. Choose one such sequence $(r_n)$. Now it suffices to show that $r_n \to \infty$, because from this it follows that $f_m(x) = 1$ for all $m$ even though $x \notin F$.

Now if $(r_n)$ was bounded, after some index $n_0$ the "route" of $x$, defined by the sequence $k(x,j)$, would differ from the "routes" of all $y \in F$ defined by the sequences $k(y,j)$ after $j \geq n_0$. This then means that for all $j \geq n_0$ and $y \in F$ we have $$|x-y| \geq d(\{x\}, I_{j, k(y,j)}) > 0\,,$$ since $x \notin I_{j, k(y,j)}$ and the distance between different intervals $I_{j,a}$ and $I_{j,b}$ is always positive. Now this inequality means that $x$ is not a limit point of $F$. This is a contradiction and thus $r_n \to \infty$. So $\lim_{m \to \infty} f_m(x) = 1 \neq \chi_{F}(x)$. This means that the pointwise convergence $f_m \to \chi_F$ fails at every limit point of $F$ that is not contained in $F$.


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