Prove $ \int_{cX} \frac{dt}{t} = \int_{X} \frac{dt}{t}$ for every Lebesgue measurable set $X$

Question

Let $c>0$. Let $X \subseteq (0,\infty)$ be a Lebesgue measurable set. Define $$ cX := \{ cx \mid x \in X \}. $$ Then $$ \int_{cX} \frac{dt}{t} = \int_{X} \frac{dt}{t}$$

Now I can prove this for $X$ an interval and, thus, any set generated by set operations on intervals. It is simply by using the Fundamental Theorem of Calculus and natural log $\ln$. But I'm not sure how to approach for general Lebesgue measurable set.

Answer 1

(I dislike the title, which looks like an assignment.)

Hint:

Since you know this for intervals, use an approximation argument of step functions for the functions $x\mapsto \chi_X(x)\cdot\frac{1}{x}$ and $x\mapsto \chi_{cX}(x)\cdot\frac{1}{x}$. Where $\chi_A$ denotes the characteristic function on $A$.

Answer 2

Suppose $m$ and $n$ are non-negative measures and $c$ is a positivie number and $n=m/c$. Can you show that $$ \int_A f\,dm = \int_A (c f)\,dn\text{ ?} $$ If you can, let $A=cX$, $m=$ Lebesgue measure, $f(t)=1/t$. Then find a one-to-one correspondence between $A=cX$ and $X$ such that the value of $1/t$ for $t\in X$ is the same as the value of $cf(t)$ for $t\in A$, and think about that.