$A$ is a borel subset of $\mathbb{R}^2$ and we know that $\lambda_2(A) = \pi$. Prove that: $\int_A (x^2 + y^2) \ d \lambda_2(x,y) \geq \frac{\pi}{2}$

Question

We assume that $A$ is a borel subset of $\mathbb{R}^2$ and that $\lambda_2(A) = \pi$. Prove that:

$$\int_A (x^2 + y^2) \ d \lambda_2(x,y) \geq \frac{\pi}{2}$$

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I see that it might be useful to use polar coordinate system. That way we would get that:

$$x = rcos(\theta) \ , \ y = rsin(\theta)$$

And we know that $\lambda_2(A) = \pi$, so we would get that:

$$\int_{\theta} \int_r \ r^2 \cdot \pi \ dr \ d\theta = \pi \int_{\theta} \int_r \ r^2 \ dr \ d\theta$$

But how to define / how to set the limits of integration so that: $$\int_{\theta} \int_r \ r^2 \ dr \ d\theta \geq \frac{1}{2}$$

Why that would be the case?

Answer 1

The following lemma says roughly that your integral will be larger if the set is ‘further out’.

Lemma

Suppose $f:\Bbb{R}^n\to\Bbb{R}$ is a radially-increasing function. If $E,F\subset\Bbb{R}^n$ are measurable sets such that $\lambda(E)=\lambda (F)$, and for some $R\geq 0$, we have $E\subset B_R(0)$ and $F\subset B_R(0)^c$, then \begin{align} \int_Ef\,d\lambda\leq\int_Ff\,d\lambda. \end{align}

The proof is pretty simple. The hypothesis means that we can write $f(x)=\phi(\|x\|)$ for some (weakly) increasing function $\phi:[0,\infty)\to\Bbb{R}$. So, because $\phi$ is increasing, we have \begin{align} \int_Ef\,d\lambda&\leq \int_E\phi(R)\,d\lambda=\phi(R)\cdot \lambda(E)=\phi(R)\cdot \lambda(F)=\int_F\phi(R)\,d\lambda\leq\int_Ff\,d\lambda. \end{align}

We have the following simple corollary:

Corollary

Let $f:\Bbb{R}^n\to\Bbb{R}$ be a radially-increasing function, and $A\subset\Bbb{R}^n$ a finite-measure set, and $R\in[0,\infty)$ be the unique value such that $\lambda(B_R(0))=\lambda(A)$. Then, \begin{align} \int_{B_R(0)}f\,d\lambda&\leq\int_Af\,d\lambda. \end{align}

To prove this, let $A_0:= A\cap B_R(0)$, let $E= B_R(0)\setminus A_0$ and $F= A\setminus A_0$. Then, we have $E\subset B_R(0)$ and $F\subset B_R(0)^c$, and also \begin{align} \lambda(E)&=\lambda(B_R(0))-\lambda(A_0)=\lambda(A)-\lambda(A_0)=\lambda(F). \end{align} So, with the help of the lemma, we have \begin{align} \int_{B_R(0)}f\,d\lambda&=\int_{A_0}f\,d\lambda+\int_Ef\,d\lambda\leq\int_{A_0}f+\int_Ff\,d\lambda=\int_Af\,d\lambda. \end{align}

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Now, apply the corollary with $f(x,y)=x^2+y^2$ and $R=1$. The left integral is exactly $\frac{\pi}{2}$.

So, as far as motivation goes, I simply noted that a direct calculation using polar coordinates gives $\int_{B_1(0)}(x^2+y^2)\,d\lambda=\frac{\pi}{2}$. So, because $\lambda(A)=\lambda(B_1(0))=\pi$, this means that the portion of the integral coming from the part of $A$ lying outside the unit ball must contribute atleast as much as the part that lies “inside”. This whole thing is just a matter of formalizing the intuition that if you have the same volume, and one set is entirely further out than the other, then the integral over the further set will be larger.