General Integration and Measure by Alan J. Weir - Exercise 4 - Section 9.2

Question

This is an exercise of Weir's book which can be read here.

$\textbf{4.}$ If the Lebesgue-Stieltjes measure $\mu_F$ is invariant under translations, show that $\mu_F$ is a constant multiple of Lebesgue measure. (In the one-dimensional case, show that $F$ is continuous and if $F(0) = 0$, satisfies $F(a+b) = F(a) + F(b)$ for all $a, b \in \mathbb{R}$).

I put what I thought below.

I will follow the hint given by the author. Since $F$ is continuous on right in a point $x_0 \in \mathbb{R}$, we have that, given $\eta := 2 \varepsilon > 0$, exist $\delta > 0$ such that

$$x \in B(x_0, \delta) \Longrightarrow F(x) \in [F(x_0), \ F(x_0) + \eta).$$

Since $\mu_F$ is invariant under translation, we have, for the same $\delta > 0$ and translating by $\frac{\eta}{2}$,

$$x \in B(x_0, \delta) \Longrightarrow F(x) \in [F(x_0) - \frac{\eta}{2}, \ F(x_0) + \frac{\eta}{2} ).$$

Thus, $x \in B(x_0, \delta) \Longrightarrow F(x) \in ( F(x_0) - \varepsilon, \ F(x_0) + \varepsilon )$, therefore $F$ is continuous at $x_0$.

We can suppose without loss of generality that $F(0) = 0$ (if $F(0) = a \neq 0$, then it's just consider $G(x) := F(x) - a$).

Given $a,b \in \mathbb{R}$, observe that

$$F(a + b) - F(a) = F(b) - F(0)$$

considering a translation by $a$, then

$$F(a + b) = F(a) + F(b).$$

I'm stuck here. Can anyone give me a direction in order to conclude that $\mu_F$ is a multiple of Lebesgue measure? The special case when $F$ is defined on $\mathbb{R}^k$ for $k = 1$ helps me to prove the general case (for $k > 1$)?

Answer 1

Decompose $I=[0,1]$ in disjoint intervals, $A_i, i=1, \ldots, n$ of same Lebesgue length $Leb(A_i)=1/n$. So by translation invariance, $\alpha=\mu_F([0,1])=n\times \mu_F(A_i)$, therefore $\mu_F(A_i)=\alpha \times Leb(A_i).$ Then we have, by tranlation invariance, $\mu_F$ and $\alpha \times Leb $ coincide in the subintervals of $I=[0,1]$, by the extension theorem they must coincide in the borelians of $[0,1]$. By translation invariance we extend this conclusion to the real line.

The general case is similar.

Answer 2

If $\mu_F$ is translation invariant, then for real $a<b$ and real $c$, $$ F(b)-F(a)=\mu_F\left((a,b]\right)=\mu_F\left((a+c,b+c]\right)=F(b+c)-F(a+c). $$ Assuming, as you have shown we may, that $F(0)=0$, the above implies (take $a=0$) that $F(b+c)=F(b)+F(c)$, at least for $b>0$. The case $b<0$ can be handled similarly. Thus $F$ is additive and continuous from the right. The only such functions are of the form $F(x)=dx$ for real $d$. This in turn implies that $\mu_F$ is $d$ times Lebesgue measure.