Show that $\int_{\mathbb{R}^n} f_t d \lambda_n = \frac{1}{|t|^n} \int_{\mathbb{R}^n} f d \lambda^n$ Suppose $f: \mathbb{R}^n \to \mathbb{R}$ is $\mathcal{B}^n$-measurable. For $t \in \mathbb{R} \setminus \{ 0 \}$, define $f_t(x) = f(tx)$.
Prove that if $\int_{\mathbb{R}^n} f d \lambda_n$ is defined then $\int_{\mathbb{R}^n} f_t d \lambda_n = \frac{1}{|t|^n} \int_{\mathbb{R}^n} f d \lambda_n$. Here $\lambda_n$ is the Lebesgure measure on $\mathbb{R}^n$.
I tried to use Tonelli and Fubini's theorems, but for that I need $f \geq 0$ or $\int_{\mathbb{R}^n} |f| d \lambda_n < \infty$, none of which are given. So I'm kinda stuck.
 A: Hint
For the kind of exercise, the proof always goes the same. Prove it first for simple function, then for positive function using approximation by increasing simple function, and then for measurable function.

*

*Step 1 : For $f(x)=\sum_{i=1}^na_i\boldsymbol 1_{A_n}(x)$ being simple, then

\begin{align*}
\int f(tx)\lambda _n(\mathrm d x)&=\sum_{i=1 }^na_i\int\boldsymbol 1_{A_i}(tx)\lambda _n(\mathrm d x)\\
&=\sum_{i=1}^na_i\int \boldsymbol 1_{t^{-1}A_n}(x)\,\mathrm d x\\
&=\sum_{i=1}^na_im(t^{-1}A_i)\\
&=\frac{1}{|t|^n} \sum_{i=1}^na_im(A_i)\\
&=\frac{1}{|t|^n}\int f(x)\lambda _n(\mathrm d x).
\end{align*}

*

*Step 2 If $f\geq 0$, there is a sequence of simple function $(\varphi _n)$ s.t. $\varphi _n\nearrow f$. I let you conclude using Monotone Convergence Theorem.


*Step 3 Take $f^+(x):= f(x)\vee 0$ and $f^-(x)=-(f(x)\wedge 0)$ (where $a\vee b$ denote the maximum of $a$ and $b$ and $a\wedge b$ denotes the minimum). Then $f^\pm \geq 0$ and $f=f^+-f^-$. You can then conclude using step $2$.
A: Let $g:\mathbb{R}^n\to \mathbb{R}^n$ such that $g(x)=tx$, $t \neq 0$. Then we use the results on image measures to assert that:
$$\int_{\mathbb{R}^n} f(tx)\lambda^n(dx)=\int_{\mathbb{R}^n} f(g(x))\lambda^n(dx)=\int_{\mathbb{R}^n} f(y)\lambda^n_g(dy)$$
where
$$\lambda_g^n(B)=\lambda^n(g^{-1}(B))=\lambda^n(t^{-1}B)=|t|^{-n}\lambda^n(B) \ \ \ \ \ \ \ B \in \mathcal{B}(\mathbb{R}^n)$$
thus
$$\int_{\mathbb{R}^n} f(tx)\lambda^n(dx)=|t|^{-n}\int_{\mathbb{R}^n} f(y)\lambda^n(dy)$$
