I'm studying for my exam in measure and integration theory and we got some exercises that we can do for preparation and I'm stuck on this one.

Every function $f:\Bbb R \rightarrow \Bbb R $ (from reals to reals) with the property that $\displaystyle\lim _{ h\to 0^{+}}{ f(x+h) } $ exists for all $x\in \Bbb R $ is Borel measurable.

I just proved that the pointwise limit of measurable functions is again measurable. So I thought about ways we could represent our $f$ as a limit of (Borel) measurable functions but I dont think it is in general possible to find such a sequence for an arbitrary $f$.

Further I feel like i don't understand the condition that $$\lim _{ h\to 0^+ }{ f(x+h) } $$ exists correctly. Is it possible to derive some form of continuity with this property? Could someone shed some light on this problem for me, thanks for any help.

  • $\begingroup$ Off-hand I don't know how to prove such a function is Borel measurable, but I KNOW that such a function has at most countably many discontinuities (follows from the much stronger results described here), and thus is actually a Baire one function (pointwise limit of continuous functions; the first level of the $\omega_1$-length hierarchy of Borel measurable functions), in fact better than Baire one because a Baire one function can be discontinuous on a $c$-dense set. $\endgroup$ – Dave L. Renfro Jan 26 at 20:13

If we set $$\tilde{f}(x) := \lim_{h \downarrow 0} f(x+h),$$ then $\tilde{f}$ is (by assumption) well-defined. Moreover, straight-forward computations show that $\tilde{f}$ is right-continuous (see Lemma 3 below) and, hence, Borel measurable. If we can prove that the set

$$J:=\{x \in \mathbb{R}; \tilde{f}(x) \neq f(x)\} \tag{1}$$

is countable, then $g:=f-\tilde{f}$ is Borel measurable (see e.g. this proof). Consequently, $f=g+\tilde{f}$ is Borel measurable as sum of Borel measurable functions.

To prove that $J$ is countable we proceed as follows: For $x \in \mathbb{R}$ define the oscillations of $f$ at $x$ by

$$\omega(x) := \inf_{r>0} \omega_r(x) := \inf_{r>0} \left( \sup_{z \in B(x,r)} f(z) - \inf_{z \in B(x,r)} f(z) \right).$$

It is not difficult to see that $$\{\omega=0\} = \{x \in \mathbb{R}; \text{$x$ is a continuity point of $f$}\} \tag{2}$$

and therefore $J \subseteq \{\omega \neq 0\}$. Consequently, we are done if we can show that $\{\omega \neq 0\}$ is countable.

Lemma 1: For any $x \in \mathbb{R}$ and $n \in \mathbb{N}$ there exists $\delta(x)>0$ such that $\omega(y) \leq 1/n$ for all $y \in (x,x+\delta(x))$.

Proof: Fix $x \in \mathbb{R}$ and $n \in \mathbb{N}$. Since the limit $\tilde{f}(x) = \lim_{h \downarrow 0} f(x+h)$ exists, we have

$$\sup_{z \in (x,x+h)} f(z) \xrightarrow[]{h \to 0} \tilde{f}(x) \qquad \inf_{z \in (x,x+h)} f(z) \xrightarrow[]{h \to 0} \tilde{f}(x),$$

and so

$$\lim_{h \to 0} \left| \sup_{z \in (x,x+h)} f(z) - \inf_{z \in (x,x+h)} f(z) \right|=0;$$

in particular we can choose $\delta>0$ such that

$$\left| \sup_{z \in (x,x+\delta)} f(z) - \inf_{z \in (x,x+\delta)} f(z) \right| \leq \frac{1}{n}. \tag{3}$$

Now let $y \in (x,x+\delta)$. If we set $r := \min\{|y-x|,|y-(x+\delta)|\}$ then $B(y,r) \subseteq (x,x+\delta)$. In particular, we have

$$\sup_{z \in B(y,r)} f(z) \leq \sup_{z \in (x,x+\delta)} f(z) \qquad \inf_{z \in B(y,r)} f(z) \geq \inf_{z \in (x,x+\delta)} f(z) \tag{4}$$

which implies

$$\inf_{z \in (x,x+\delta)} f(z) \leq \inf_{z \in B(y,r)} f(z) \leq \sup_{z \in B(y,r)} f(z) \leq \sup_{z \in (x,x+\delta)} f(z).$$

On the other hand, we know from $(3)$ that

$$\sup_{z \in (x,x+\delta)} f(z) \leq \inf_{z \in (x,x+\delta)} f(z) + \frac{1}{n}.$$

Combining the two chains of inequalities we conclude that

$$\sup_{z \in B(y,r)} f(z) - \inf_{z \in B(y,r)} f(z) \leq \frac{1}{n},$$

i.e. $\omega_r(y) \leq 1/n$. In particular, $\omega(y) \leq 1/n$ which finishes the proof of the Lemma.

Lemma 2: $\{\omega \neq 0\}$ is countable.

Proof: Clearly, it suffices to show that $\{\omega > 1/n\}$ is countable for each $n \in \mathbb{N}$. For fixed $n \in \mathbb{N}$ denote by $\delta(x)$ the constant from the previous lemma. For each fixed $k \in \mathbb{N}$ and $N \in \mathbb{N}$ the set

$$B_{k,N} := \{x \in [-N,N] \cap \{\omega>1/n\}; \delta(x) \geq 1/k\}$$

is finite. Indeed: By the previous lemma, the distance between any two points in $B_{k,N}$ is at least $1/k$ and since the length of the interval $[-N,N]$ is $2N$, there can exist at most $2Nk+1$ points in $B_{k,N}$. This implies that

$$\{x \in \{\omega>1/n\}; \delta(x) \geq 1/k\} = \bigcup_{N \in \mathbb{N}} B_{k,N}$$

is countable which, in turn, implies that

$$\{\omega>1/n\} = \bigcup_{k \in \mathbb{N}} \{x \in \{\omega>1/n\}; \delta(x) \geq 1/k\}$$

is countable.

Edit: Following the comment to my answer, I add a proof for the right-continuity of $\tilde{f}$.

Lemma 3: $\tilde{f}$ is right-continuous.

Proof: Since $\tilde{f}(y) = \lim_{h \downarrow 0} f(y+h)$ we clearly have

$$\inf_{z \in (y,y+r)} f(z) \leq \tilde{f}(y) \leq \sup_{z \in (y,y+r)} f(z) \tag{5}$$

for any $y \in \mathbb{R}$ and $r>0$. For fixed $x \in \mathbb{R}$ and $\epsilon=1/n$ let $\delta=\delta(x)>0$ be as in (the proof of) Lemma 1. Using (5) for $y=x$ we find that

$$\inf_{z \in (x,x+\delta)} f(z) \leq \tilde{f}(x) \leq \sup_{z \in (x,x+\delta)} f(z).$$

On the other hand it follows from (4) and (5) that

$$\inf_{z \in (x,x+\delta)} f(z) \leq \inf_{z \in B(y,r)} f(z) \leq \tilde{f}(y) \leq \sup_{z \in B(y,r)} f(z) \leq \sup_{z \in (x,x+\delta)} f(z)$$

for any $y \in (x,x+\delta)$ where $r:=\min\{|y-x|,y-(x+\delta)|\}$. Combining both inequalities and using (3) we get

$$|\tilde{f}(x)-\tilde{f}(y)| \leq \left| \sup_{z \in (x,x+\delta)} f(z) - \inf_{z \in (x,x+\delta)} f(z) \right| \leq \frac{1}{n}$$

for all $y \in (x,x+\delta)$ which finishes the proof of the right-continuity of $\tilde{f}$.

  • $\begingroup$ great reasoning! just to make sure that i understand correctly the borel measuablility of g:=f−f follows from the fact that every countable set in R is Borel measurable? $\endgroup$ – MasterPI Jan 27 at 12:12
  • $\begingroup$ @MasterPI Yes, essentially. We can write $g$ in the form $$g(x) = \sum_{j=1}^{\infty} c_j 1_{\{x_j\}}(x)$$ where $(x_j)_j$ is an enumeration of $J$; using the fact that every countable subset of $\mathbb{R}$ is Borel measurable, it can be easily check from the definition of (Borel) measurability that $g$ is Borel measurable. $\endgroup$ – saz Jan 27 at 12:26
  • $\begingroup$ That $\bar{f}$ is Borel also requires a proof. Note that for any open set $U$, $\bar{f}^{-1}(U)$ is an open set with respect to the "lower limit topology". The "lower limit topology" is strictly larger than the usual topology on $\mathbb{R}$. It is true that the $\sigma$-algebra generated by these two topologies are the same but the proof is hard and tricky. $\endgroup$ – Danny Pak-Keung Chan Jan 28 at 0:37
  • $\begingroup$ @DannyPak-KeungChan It requires a proof, yes, but it's not that difficult. As I indicated in my answer, I would use the right-continuity of $\tilde{f}$ to conclude that $\tilde{f}$ is Borel measurable. Showing that $\tilde{f}$ is right-continuous is a bit tedious but not difficult (I've now added the proof, see Lemma 3) and that right-continuous functions are Borel measurable is a well-known fact which is also not difficult to prove (e.g. via an approximation argument, as in this answer). $\endgroup$ – saz Jan 28 at 5:19

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