Show that $\int f_2(x)dx=\frac{1}{2}\int f(x)dx$ if $f_2(x)=f(2x)$.

Question

Let $f$ an integrable function on $\mathbb R$ and $f_2(x)=f(2x)$. I have to show that $$\int f_2(x)dx=\frac{1}{2}\int f(x)dx.$$ I have the idea but problem to write it properly.

For simple function: Let $f(x)=\sum_{i=1}^n a_i\boldsymbol 1_{F_i}(x)$ where $F_i$ are measurable. I have that $f_2(x)=\sum_{i=1}^n a_i\boldsymbol 1_{F_i}(2x)$. Then $$\int f=\sum_{i=1}^n a_im(F_i).$$ I'm sure that $$\int f_2(x)=\sum_{i=1}^n2a_im(F_i)$$ but I have problem to convert $1_{F_i}(2x)$ in something like $1_{G_i}(x)$ where $m(G_i)=2m(F_i)$. Thank you for helping.

Answer 1

Note that

$$1_{F_i}(2x) = \begin{cases} 1 & \text{if }2x \in F_i \\ 0 & \text{otherwise.}\end{cases}$$

This mean

$$1_{F_i}(2x) = \begin{cases} 1 & \text{if }x \in F_i/2 \\ 0 & \text{otherwise.}\end{cases} = 1_{F_i/2} (x)$$

where $F_i/2 = \{x/2 : x\in F_i\}$. Note that you want $m(F_i/2) = \frac 12 m(F_i)$.