0

Let $D$ be a Lebesgue measurable set and $y\in \mathbb{R}$. If we assume that the Lebesgue measure $m(D\Delta(D+y))=0$. Here $A\Delta B=(A\setminus B)\cup (B\setminus A)$. How do we conclude that $$ \chi_D(x)=\chi_{D+y}(x). \, a.e. $$

I try to rewrite it as integral $$ 0=m(D\Delta(D+y))=\int_{D\Delta (D+y)}dm(x) $$ But I am stuck on how to show that $$ \int_{D\Delta (D+y)}dm(x)=\int_\mathbb{R} |\chi_D(x)-\chi_{D+y}(x)|dm(x) \ge 0? $$

Moreover, after we have $$ \int_\mathbb{R} |\chi_D(x)-\chi_{D+y}(x)|dm(x)=0 $$

How do we get $$ \chi_D(x)=\chi_{D+y}(x). \, a.e.? $$

Hermi
  • 1,480
  • What exactly do you mean by the $\Delta$ operator here? – o.spectrum Feb 05 '24 at 22:55
  • The fact that the Lebesgue integral evaluates to zero implies that the integrated function is zero a.e. (there's actually a proof here at Math SE, so you can find it if you want). This is exactly how you get from the last integral to the desired equation. – o.spectrum Feb 05 '24 at 22:59
  • And by the way, what is the nature of $D$? Is it countable/uncountable? – o.spectrum Feb 05 '24 at 23:09
  • 1
    $f \ge 0, \int fd\mu=0$ implies $f=0$ a.e. This is a very basic result in measure theory. – geetha290krm Feb 05 '24 at 23:10
  • @o.spectrum Here is the set difference $A\Delta B=(A\setminus B)\cup (B\setminus A)$ – Hermi Feb 05 '24 at 23:44
  • @o.spectrum I think $D$ can be uncountable... Does it matter? – Hermi Feb 05 '24 at 23:45
  • @Hermi it quite didn't, just I wanted to understand what $D+y$ means, and how to write it in terms of sets, so that was kind-of more of an idle question :) – o.spectrum Feb 06 '24 at 00:04
  • @o.spectrum Just fix $y$, and $E+y:={e+y: e\in E}$. – Hermi Feb 06 '24 at 04:43

1 Answers1

0

Let us assume that by the $\Delta$ operator you mean the symmetric difference and that for a given set $D=\{d_1, d_2, \ldots,d_n,\ldots\}$ by $D+y$ you mean $\{d_1+y, d_2+y,\ldots,d_n+y,\ldots\}$. Let us also assume that $m(D)<+\infty$ and $m(D+y)<+\infty$.

Let us rewrite your integral a bit: \begin{align} 0&= \int_{D\Delta D+y}\;m(dx)\\ &=\int_{(D\setminus D+y)\cup (D+y \setminus D)}\;m(dx)\\ &=\int_{\mathbb{R}}\chi_{(D\setminus D+y)\cup (D+y \setminus D)}(x)\;m(dx)\\ &=\int_{\mathbb{R}}\chi_{D\setminus D+y}(x) \oplus\chi_{D+y\setminus D}(x)\;m(dx)\\ &=\int_{\mathbb{R}}|\chi_{D\setminus D+y}(x)| \oplus|\chi_{D+y\setminus D}(x)|\;m(dx)\\ &= \int_{\mathbb{R}}|\chi_{D\setminus D+y}(x) -\chi_{D+y\setminus D}(x)|\;m(dx)\;\text{ (by the properties of xor)}\\ &= \int_{\mathbb{R}}|(\chi_{D\setminus D+y}(x)+\chi_{D\cap D+y}(x)) -(\chi_{D+y\setminus D}(x)+\chi_{D+y\cap D}(x))|\;m(dx)\\ &= \int_{\mathbb{R}}|(\chi_{(D\cap (D+y)^c)\cup(D\cap D+y)}(x)) -(\chi_{(D+y\cap D^c)\cup(D+y\cap D)}(x))|\;m(dx)\\ &=\int_{\mathbb{R}}|\chi_D(x)-\chi_{D+y}(x)|\;m(dx) \end{align} Since we have that by given the measure over the symmetric difference is zero, we have that the last integral evaluates exactly to zero as well. After that you can apply the fact mentioned by @geetha290krm and get that $|\chi_D(x)-\chi_{D+y}(x)|=0$ a.e., and further that $$ \chi_D(x)=\chi_{D+y}(x)\; \text{ a.e.} $$

o.spectrum
  • 1,170
  • 8
  • 19
  • Thanks. But I think $|\chi_D(x)-\chi_{D+y}(x)|=\chi_{D\Delta (D+y)}(x)$, right? – Hermi Feb 06 '24 at 02:28
  • Also, can we say that $\chi_D(x)=\chi_D(x+y)$ a.e.? – Hermi Feb 06 '24 at 02:29
  • No, in fact you cannot make such a direct equality there in general. It's rather $\chi_{D\Delta D+y}(x)=\chi_{D}(x)\oplus \chi_{D+y}(x)$ in general case scenario. This is only the assumption of yours that $m(D\Delta D+y)=0$ that allows to state us that $\chi_{D\Delta D+y}(x)=|\chi_{D}(x)-\chi_{D+y}(x)|$ in this very case. – o.spectrum Feb 06 '24 at 02:33
  • I'm not sure what you mean by $\chi_D(x+y)$, if $y$ in this context is just some scalar then yes, because in fact it's not the variable, we evaluate the indicator function over, that matters but rather its domain. For me at least, $\chi_D(x)=\chi_D(x+y)$ a.e. doesn't really make any practical sense, since, well, $\chi_D$ is just the very same indicator for both LHS and RHS here. – o.spectrum Feb 06 '24 at 02:36
  • Ok, because I try to understand the answer in this question:https://math.stackexchange.com/questions/103473/revisiting-a-lebesgue-measure-question-involving-a-dense-subset-of-r-translates?noredirect=1&lq=1 – Hermi Feb 06 '24 at 04:36
  • But I discuss all cases for $x\in A\cap B$, and $ x\in A$ but $x\notin B$,...I can show that $|\chi A-\chi_B|=\chi{A\Delta B}$. I am not sure why do you say they are not equal? Can you give me one case that they are not equal? – Hermi Feb 06 '24 at 04:39
  • See here https://math.stackexchange.com/questions/3457402/symmetric-difference-characteristic-function-lebesgue-norm/3457442#3457442 – Hermi Feb 06 '24 at 04:42
  • Oh, in fact you're right. Just I seem to have made a slip myself, as I was actually relating to measures first, not the indicator functions. Instead of a "$+$" under the integral, we must actually have "$\oplus$" (the xor), and then the direct equality can be drawn. – o.spectrum Feb 06 '24 at 10:13