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

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

  • 1
    $\begingroup$ why the downvote? $\endgroup$
    – vidyarthi
    Apr 2, 2019 at 12:01
  • 1
    $\begingroup$ 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 $\prod_{i=1}^d [a_i, b_i[$. $\endgroup$
    – bangs
    Apr 2, 2019 at 12:16

1 Answer 1


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.

Let me know if something is unclear.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .