This question comes from the exercises of Stein and Shakarchi's Real analysis Ex. 5.14.


$$ j_n(x)= \begin{cases} 0& \text{if } x< x_n\\ \theta_n & \text{if } x=x_n\\ 1 &\text{if } x>x_n \end{cases} $$

where $\{x_n\}$ is a sequence of real number and $0\le\theta_n\le 1$.

Further define


where $\alpha_n>0$ and $\sum\alpha_n$ converges.

Prove that for any $\epsilon$, the set of $x$ satisfying $$\limsup_{h\to0}\frac{J(x+h)-J(x)}{h}>\epsilon$$

is measurable.

So far what I can prove is that the sum in $J(x)$ can converge absolutely. And I know the limit of measurable functions is measurable. However, Since $J$ is not continuous, I cannot assume $h\to0$ as $\frac1n;n\to\infty$, then I cannot figure out whether the above set is measurable.

Edit: There is a hint in the book: consider $$F^N_{k,m}(x)=\sup_{1/k\le|h|\le1/m}\left|\frac{J_N(x+h)-J_N(x)}{h}\right|$$ where $J_N$ is the $N^{\text{th}}$ partial sum of $J$

Show that $F^N_{k,m}$ is measurable (which I couldn't tell why) and so does $$\lim_{m\to\infty}\lim_{k\to\infty}\lim_{N\to\infty}F^N_{k,m}(x)$$

(Which I also cannot relate it to the original set because I am confused of the multiple limit and supremum)

  • $\begingroup$ Just to be clear, the exercise asks you to show that the map $x\mapsto \limsup_{h\to0}{J(x+h)-J(x)\over h}$ is measurable. The measurability of the set follows from this. $\endgroup$
    – user940
    Aug 8, 2015 at 22:19

2 Answers 2


For brevity, write

$$\Delta(x,h) := \frac{J(x+h)-J(x)}{h} \qquad \text{and} \qquad \Delta_n(x,h) := \frac{J_n(x+h)-J_n(x)}{h} $$

Step 1: By definition of limsup, we have

$$\begin{align*} \limsup_{h \to 0} \Delta(x,h) > \epsilon &\iff \forall m \in \mathbb{N} \, \exists h \in \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \{0\}: \Delta(x,h) > \epsilon \\ &\iff \forall m \in \mathbb{N} \, \exists k \geq m \, \exists h \in \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \left(- \frac{1}{k}, \frac{1}{k} \right): \Delta(x,h)>\epsilon \\ &\iff \forall m \in \mathbb{N} \, \exists k \geq m: \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon. \end{align*}$$

This shows that

$$\{x; \limsup_{h \to 0} \Delta(x,h)>\epsilon\} = \bigcap_{m \in \mathbb{N}} \bigcup_{k \geq m} \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon \bigg\}.$$

Consequently, it suffices to prove that for each fixed $k,m \in \mathbb{N}$ and $\epsilon>0$ the set

$$A := A_{k,m,\epsilon} := \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon \bigg\} \tag{1}$$

is measurable.

Step 2: We claim that $$A = \bigcup_{\ell \in \mathbb{N}} \bigcup_{N \in \mathbb{N}} \bigcap_{n \geq N} \bigg\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon + \frac{1}{\ell} \bigg\}. \tag{2}$$

For brevity of notation, we set $$I := I_{k,m} := \left[- \frac{1}{m}, \frac{1}{m} \right] \backslash \left(- \frac{1}{k}, \frac{1}{k} \right).$$

Proof: Let $x \in A$, then $\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h)>\epsilon$. By the definition of $\sup$, we can find $h \in I$ such that $\Delta(x,h)>\epsilon$. In particular, there exists $l \in \mathbb{N}$ such that $\Delta(x,h) > \epsilon + \frac{2}{\ell}$. As $\Delta(x,h) = \lim_{n \to \infty} \Delta_n(x,h)$, we have $\Delta_n(x,h) \geq \epsilon+ \frac{1}{\ell}$ for all $n$ sufficiently large. This proves "$\subseteq$".

Now let $x$ be an element of the right-hand side of $(2)$. Then there exist $\ell ,N \in \mathbb{N}$ such that $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon + \frac{1}{\ell}$$ for all $n \geq N$. By the definition of $\sup$, we can find $h(n) \in I$ such that $$\Delta_n(x,h(n))> \epsilon + \frac{1}{\ell}. \tag{3}$$ Since $I$ is compact, there exists a convergent subsequence $h_n \to h \in I$. Choose $n=n(x) \geq N$ sufficiently large such that $$|J_n(x)-J(x)| \leq \frac{1}{2k \ell}.$$ This implies $$\frac{|J_n(x)-J(x)|}{h} \leq \frac{1}{2\ell}$$ for all $h \in I$; hence in particular for $h=h_n$. Combining this with $(3)$, we get

$$\begin{align*} \frac{J_n(x+h_n)-J(x)}{h_n} &= \frac{J_n(x+h_n)-J_n(x)}{h_n} + \frac{J_n(x)-J(x)}{h_n}\\ &\geq \epsilon + \frac{1}{\ell} - \frac{1}{2\ell} = \epsilon + \frac{1}{2\ell} > \epsilon. \end{align*} \tag{4}$$

Finally, we note that $J_n(y) \leq J(y)$ implies

$$\epsilon \stackrel{(4)}{<} \frac{J_n(x+h_n)-J(x)}{h_n} \leq \frac{J(x+h_n)-J(x)}{h_n} = \Delta(x,h_n).$$

Consequently, $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta(x,h) \geq \Delta(x,h_n) > \epsilon,$$ i.e. $x \in A$.

Step 3: Recall that we actually want to show that $A$ defined in $(1)$ is measurable. By the second part, it suffices to show that sets of the form $$ B := B_{N,n,k,m,\epsilon} := \left\{x; \sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h)> \epsilon \right\}$$ are measurable.

To show this, we use the following elementary result:

Lemma: Let $f: (a,b) \to \mathbb{R}$ be of the form $f = \frac{f_1}{f_2}$ where $f_1$ is increasing and $f_2 \neq 0$ continuous. Then $$\limsup_{\mathbb{Q} \ni q \downarrow x} f(q) \geq f(x)$$ for any $x \in (a,b)$.

Since the mapping $h \mapsto J_n(x+h)-J_n(x)$ is increasing, we may apply the previous result to obtain $$\sup_{\frac{1}{k} \leq |h| \leq \frac{1}{m}} \Delta_n(x,h) = \sup_{(\mathbb{Q} \cap I) \cup \{\frac{1}{m}\}} \Delta_n(x,h) \tag{5}$$ where $I$ is defined as in step 2. (The inequality "$\geq$" holds in any case, the lemma above gives "$\leq$".) Since $\Delta_n(\cdot,h)$ is measurable, this proves that the left-hand side of $(5)$ is measurable and so is $B$.

  • $\begingroup$ (If anyone else comes up with a better idea, I'll remove my answer. Might well be that my proof is much too complicated.) $\endgroup$
    – saz
    Aug 8, 2015 at 10:36
  • $\begingroup$ Wow! It is a great proof. I just have one question, which is why we can assume that $x+h\ne x_k$ in your second part? $\endgroup$
    – Y.H. Chan
    Aug 8, 2015 at 13:35
  • $\begingroup$ @UnemChan I have changed the second part; there was some mistake in it. Should be fine now... $\endgroup$
    – saz
    Aug 8, 2015 at 15:26
  • $\begingroup$ @saz It seems to me that the set $I$ in equation (3) might depend on $x$. $\endgroup$
    – user940
    Aug 8, 2015 at 22:32
  • $\begingroup$ @ByronSchmuland I don't think so, but to get rid of this hand-waving argumentation, I have also changed this part of my answer (...which is now pretty close to yours, but, in fact, I had already thought of this before - I simply didn't realize that this can be used to show the measurability directly, without this approximation-stuff. +1 for your answer :)) [And thanks for correcting my typos...] $\endgroup$
    – saz
    Aug 9, 2015 at 9:49

$J$ is a non-decreasing function, so for any fixed $h\neq 0$ if $q_n\to h$ with $|q_n|\geq |h|$, then $$\lim_n {J(x+q_n)-J(x)\over q_n} \geq {J(x+h)-J(x)\over h}\geq 0.$$ (Note that the absolute value bars are not necessary.) Hence for $m\geq 1$ we have $$\sup_{q \in\mathbb{Q},\, 0<|q|<1/m}{J(x+q)-J(x)\over q}= \sup_{0<|h|<1/m} {J(x+h)-J(x)\over h},\tag1$$ where $\mathbb{Q}$ means the rational numbers. Since $\mathbb{Q}$ is a countable set, equation (1) shows that $$\Psi_{m}(x):= \sup_{0<|h|<1/m}{J(x+h)-J(x)\over h}$$ is a measurable map into $[0,\infty].$ Therefore
$$\limsup_{h\to0}{J(x+h)-J(x)\over h} =\inf_{m\geq 1} \Psi_{m}(x)$$ is also measurable into $[0,\infty].$


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