# Show that $\phi (\lambda^{d})=\lambda^{d}$

Let $$d \in \mathbb N$$ and define $$\lambda^{d}$$ as the lebesgue borel measure. Let $$\phi: \mathbb R^{d} \to \mathbb R^{d}$$ be measurable and $$\phi(x)=-x$$

Show that $$\phi (\lambda^{d})=\lambda^{d}$$

and use this to show that $$\int_{\mathbb R^{d}}f(-x)d\lambda^{d}(x)=\int_{\mathbb R^{d}}f(x)d\lambda^{d}(x)$$

In order to show that two measures $$\phi (\lambda^{d})$$ and $$\lambda^{d}$$ are equal I need to show that they are equal on on generator of $$\mathcal{B}^{d}$$.

My idea: Let $$[a,b[$$ be an arbitrary set in $$\mathbb R^{d}$$ then $$\phi(\lambda^{d})([a,b[)=\lambda^{d}(\phi^{-1}([a-b[))=\lambda^{d}(-[a-b[)=\vert-1\vert^{d}\lambda^{d}([a,b[=\lambda^{d}([a,b[)$$. Does this suffice? I feel unsure of "negating a set".

If $$\phi(\lambda^{d})=\lambda^{d}$$ holds, then

$$\int_{\mathbb R^{d}}f(x)d\lambda^{d}(x)=\int_{\mathbb R^{d}}f(x)d\phi(\lambda^{d})(x)=\int_{\phi^{-1}(\mathbb R^{d})}f\circ \phi (x)d\lambda^{d}(x)=\int_{\mathbb R^{d}}f(-x)d\lambda^{d}(x)$$

• why the downvote? Apr 2, 2019 at 12:01
• You can't say "Let $[a,b[$ be an arbtirary set", because having it written the way you do implies that it is an interval. Moreover, since you are working in $\mathbb{R}^d$, you should take an arbitrary product of $d$ intervals, and not a single interval. Third, you should not have $\lambda^d (\phi^{-1}([a-b[))$, because $[a-b[$ does not make sense. It should be $$\phi(\lambda^d) ([a,b[)= \lambda^d(\phi^{-1}([a,b[)),$$ with $[a,b[$ in both places. And $[a,b[$ should actually be $\prod_{i=1}^d [a_i, b_i[$. Apr 2, 2019 at 12:16

Two measures are equal if they are equal on every measurable set. So let $$A\subset \mathbb{R}^n$$ be any measurable set. The lebesgue measure is $$\lambda^d (X) = \int_{A} 1 dx = \int_{\mathbb{R}^d} \chi_{A}(x) dx,$$ where $$\chi_A$$ is the characteristic funcion. $$\chi_A(x)=\left\{ \begin{matrix} 1 \text{ if } x \in A \\ 0 \text{ if } x \not\in A \end{matrix} \right.$$ I assume $$\phi(\lambda^d)$$ is the pushforward measure of $$\lambda^d$$ via $$\phi$$, usually denoted as $$\phi_* \lambda^d$$, defined as $$\phi(\lambda^d) (A) = \lambda^d(\phi^{-1}(A))$$ and we get \begin{align} \phi(\lambda^d) (A) &= \int_{ \mathbb{R}^d} \chi_{\phi^{-1}(A)}(x) dx = \int_{\mathbb{R}^d} \chi_{-A}(x) dx = \int_{\mathbb{R}^d} \chi_A(-x) dx \\ &= \int_{\mathbb{R}^d} \chi_A(y) |\det d\phi(x)| dy = \int_{\mathbb{R}^d} \chi_A(x) dx = \lambda^d(A), \end{align} where we used the substitution formula.
Comments about your proof: As has been pointed out in the comments, you are in the $$d$$-dimensional setting and an interval is not an arbitrary set, (but what you mean is probably that the set of intervals make up a generator of the borel sets). Also, your formula $$\lambda^d ([a,b])=(-1)^d \lambda ^d ([b,a])$$ is really wrong, as measures are always positive. So no, your proof is not correct and it does not suffice. The last calculation is correct though.