Let $f\in L^{1}(\mathbb{R}^d), a_1,\dots,a_d>0$, and $a=(a_1,\dots,a_d)$. Define $$g(x)=f(a_1^{-1}x_1,\dots,a_d^{-1}x_d).$$ Show that $d\in L^{1}(\mathbb R^d)$ and that $$\int g=\left(\prod^{d}_{j=1}a_j\right)\int f.$$

$\textbf{My Attempt:}$ Since $f$ is integrable, it is also measurable, and hence there is an increasing sequence of simple functions $(\varphi_n)_n$, such that $\varphi_n\to f$ a.e. This implies that $\varphi_n(a^{-1}x)\to f(a^{-1}x)$ a.e, where $a^{-1}x\equiv (a_1^{-1}x_1,\dots,a_d^{-1}x_d).$ This implies that $g$ is measurable since it is a limit of measurable functions.

Let $\varphi(x)=\sum_{j=1}^{N}c_j\cdot1_{E_j}(x)$ be a simple function, where the $E_j$ are measurable sets. Then we have by dilation invariance of the Lebesgue measure and linearity of the Lebesgue integral we get for $\psi(x)=\varphi(a^{-1}x)$ $$\large\int\psi=\sum^{N}_{j=1}c_km(a^{-1}E_j)=\prod^{d}_{j=1}a_j\sum^{N}_{j=1}c_k m(E_j).$$ For a non-negative integrable function, the monotone convergence theorem comes to the rescue, since we can approximate our function by an increasing sequence of simple functions, and then the result follows from the above.

Here is the work. Let $g$ be as at the start, but non-negative for simplicity. Then we have by the remark, using the Monotone convergence theorem that $$\large\int g=\lim\limits_{n\to\infty}\int\varphi_{n}(a^{-1}x)=\prod^{d}_{j=1}a_j\lim\limits_{n\to\infty}\int\varphi_{n}=\prod^{d}_{j=1}a_j\int f.$$

Is my work above correct? Any feedback is much welcomed.

Thank you for your time.

  • 1
    $\begingroup$ $m([0,2]) = m(2[0, 1]) = 2m([0, 1]) = 2$, right? So something is shady in here. $\int \varphi_n(a^{-1}x)dx = \sum c_jm(aE_j)$, $1_E(a^{-1}x) = 1_{aE}(x)$ $\endgroup$ – Jakobian Mar 16 at 20:53
  • $\begingroup$ @Jakobian My bad, I worked out that $1_{E}(ax)=1_{a^{-1}E}(x)$ for $\mathbb{R}$ and then we extend this to $\mathbb{R}^d$ by using the product measure, since if $E=E_1\times\cdots\times E_d$, then $m(a^{-1}E)=\prod^{d}_{j=1}m(a_{j}^{-1}E_j)=\prod^{d}_{j=1}a_{j}^{-1}m(E)$. Is that right? $\endgroup$ – G the Stackman Mar 16 at 21:50
  • $\begingroup$ Yes, that's right $\endgroup$ – Jakobian Mar 16 at 22:47
  • $\begingroup$ This is not dilation invariance. It's the dilation property of the Lebesgue integral in this setting. $\endgroup$ – zhw. Mar 17 at 15:25
  • $\begingroup$ @zhw. My mistake, sorry. I corrected the error. $\endgroup$ – G the Stackman Mar 20 at 9:38

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