Measurable function and difference set

Question

I don't understand this task and how to solve it:

Let $\lambda$ denote the Lebesgue measure on ($\mathbb{R}$,$\mathfrak{B}(\mathbb{R})$). Recall that for any $E \in \mathfrak{B}(\mathbb{R})$ with $\lambda(E)>0$ the set $E-E:=\{x-y:x,y\in E\}$ contains some open intervals centered at $0$: $\exists\epsilon>0,]-\epsilon,\epsilon[ \subset E-E$

Suppose $f:\mathbb{R}\rightarrow \mathbb{R}$ is a measurable function such that $\ \forall(x,y)\in \mathbb{R}^2\ f(x+y)=f(x)+f(y)$.

1. for $k\in\mathbb{N}$ justify that the set $E_k :=\{x\in\mathbb{R}:|f(x)|<k\}$ is in $\mathfrak{B}(\mathbb{R})$ and, by observing the identity $$\bigcup\limits_{n\in\mathbb{N}}\uparrow E_k=\mathbb{R}$$ show that there exist $k\in \mathbb{N}$ and $\epsilon>0$ such that $|x|<\epsilon \implies |f(x)|<2k$.

2. Deduce that $f(x) \rightarrow 0$ as $x\rightarrow 0$.

Well first I'm not sure, what this $\uparrow$ should mean, but I guess it symbolizes just that it is an increasing union.

For the first part we see that each set $E_k$ induced an open interval of the form $(-\infty,k)$ for $k\in\mathbb{N}$ as an image and therefore $f^{-1}((-\infty,k) \subset \mathfrak{B}(\mathbb{R})$. So it's clear that $f:(\mathbb{R}$,$\mathfrak{B}(\mathbb{R}))$$\rightarrow (\mathbb{R}$,$\mathfrak{B}(\mathbb{R}))$

Now here is where I get stucked. I don't get the link to $E-E$. If I take $E_1$ I see that $|f(x)|<1$ and the max centered interval is $(-1,1)$ and for $E_k$ therefore $(-k,k)$. I thought I could somehow work with $|f(x+y)|=|f(x)+f(y)|<k$ but I really don't know what to do.

Thank you for your help

Answer 1

First of all the measurability follows from $E_k=f^{-1}(-k,k)$ more easily.

The arrow suggests that this is an increasing sequence and all we have to get from the equation is that there exists an index $k$ such that the Lebesgue measure $\lambda(E_k)>0$ (otherwise we could conclude that the Lebesgue measure of $\mathbb{R}$ is zero).

For this index $k$ we can use the theorem you cited, i.e., there exists an $\epsilon>0$ such that $]-\epsilon, \epsilon[\subseteq E_k-E_k$.

For $x\in ]-\epsilon, \epsilon[$ we thus have $x=y_1-y_2$ for some $y_1, y_2\in E_k$ (this implies $|f(y_i)|<k$).

From your assumption we have that $f(x+y_2)=f(x)+f(y_2)$, and we also have $f(x+y_2)=f(y_1)$. Therefore, we obtain $f(x)=f(y_1)-f(y_2)$.

Using the triangle inequality, we finally get

$$|f(x)|\leq |f(y_1)|+|f(y_2)|< k+k=2k$$

For the second point we first note that $2f(x)=f(x)+f(x)=f(2x)$ implies $2^nf(x)=f(2^nx)$ (prove by induction).

Then we have for $x\in ]-\frac{\epsilon}{2^n},\frac{\epsilon}{2^n}[$ that $|2^nf(x)|=|f(2^nx)|<2k$ and thus $|f(x)|<\frac{k}{2^{n-1}}$ implying the second point.