I've been looking everywhere for help (and I've haven't got very far as of yet [!]), and I unfortunately don't have much insight in terms of starting this proof. The problem-statement is given as followed:

Problem (Exact): Let $X$ be a measurable subset of $\mathbb{R}$, with finite, positive measure, $m(X)$. Let $f:X\rightarrow\mathbb{R}$ be bounded and measurable, such that $f$ obeys the relation ${\displaystyle{\bigg(\int_{X} ~f\bigg)^{2}=m(X)\int_{X}~f^{2}}}$. Prove that $f$ is almost everywhere constant.

I must apologize that I don't have much preliminary work, since I skipped over the problem and went onto working others. Any help, suggestions, recommendations, hints, tips, etc., will be GREATLY APPRECIATED! I recall that I kept getting stumped trying to bring the assumptions together to get somewhere close to a method to show $f$ is constant except on some zero set.

Also, I'm familiar with the (Lebesgue) measurability of $X$ given by either of the definitions $m^{*}(X)=\inf\big\{m(U):X\subset U,~U\text{ open}\big\}=m(X)=m_{*}(X)=\sup\big\{m(V):V\subset X,~V\text{ closed}\big\}$; or, e.g., $m^{*}(X)=\inf\bigg\{{\displaystyle{\sum\limits_{k=1}^{+\infty}|I_{k}|:X\subset\bigcup\limits_{k=1}^{+\infty}I_{k},~I_{k}\subset\mathbb{R}\text{ is open},~k\in\mathbb{N}}}\bigg\}$ and/or $X$ is measurable if for any test set $E\subset\mathbb{R}$ we have $m^{*}(E)=m^{*}(E\cap X)+m^{*}(E\cap X^{c})$.

  • 1
    $\begingroup$ This is related to the probability fact that for a random variable $X$ we have $Var(X)=E[X^2]-E[X]^2$. You might define $z=\frac{\int_X f}{m(X)}$ and then compute $\int_X (f-z)^2$. $\endgroup$ – Michael Mar 25 '16 at 18:48
  • $\begingroup$ @Michael I'll give it a shot, and thank you for the advice! I have to unfortunately admit that I have very little experience with probability. But, from what I'm gathering (if I'm correct) - the random variable statement is used to start off with, then computing the integral and coupling the result with the random variable statement is the overall way to go? $\endgroup$ – Procore Mar 25 '16 at 18:55
  • $\begingroup$ You don't need to use probability. But, a finite measure space can be converted to a probability measure simply by normalizing everything by dividing by the measure of the whole space, which is the inspiration for dividing by $m(X)$ in my comment above. $\endgroup$ – Michael Mar 25 '16 at 18:58
  • $\begingroup$ @Michael: I see what you mean (thanks again)...I'll see what I can come up with - your comment alone provides more help than hours of time I've spent pondering how to start the proof as well as checking online for help afterwards - I usually post question here as kind of a last resort in general. $\endgroup$ – Procore Mar 25 '16 at 19:01

Basically this is just the "converse" of Schwarz's inequality. One has $$\left(\int_X f\right)^2=\left(\int_X 1\cdot f\right)^2\leq \int_X 1^2\cdot \int_X f^2 =m(X)\cdot\int_X f^2$$ with equality sign at $\ \leq\ $ iff there is a $\lambda\in{\mathbb R}$ such that $\ f(x)=\lambda \>1(x)\equiv\lambda$ almost everywhere on $X$ with respect to $m$.

  • $\begingroup$ I follow this completely! I'm still new to Lebesgue and measure theory. I have to remember that the set of measurable functions from $\mathbb{R}$ to $\mathbb{R}$ is a vector space (integrable functions a subspace of this $\mathbb{R}$-vector space)!!! Thank you Christian. $\endgroup$ – Procore Mar 25 '16 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.