If $\int_{ \mathbb{R} } f^{2n} d \lambda = \int_{ \mathbb{R} } f d \lambda $ then $\lambda (f^{-1}((1,\infty)))=0$

Question

Consider the measure space $(\mathbb{R}, B( \mathbb{R}), \lambda)$, where $B( \mathbb{R}) $ denotes the Borel $\sigma$-algebra and $\lambda$ denotes the Lebesgue measure.

a) Let $f\in L^1$ be a function with values in $ [0,1]$ such that $$\int_{ \mathbb{R} } f^2 d \lambda = \int_{ \mathbb{R} } f d \lambda . $$

Show that there exists a borel set $B$ of finite measure such that $f$ is the characteristic function almost everywhere.

b) Let $f\in L^1$ be a positive function such that $\forall n \geq 1$ $$\int_{ \mathbb{R} } f^{2n} d \lambda = \int_{ \mathbb{R} } f d \lambda . $$

1) Show that $\forall \alpha >0,$ $\lambda (f^{-1}([1+ \alpha ,\infty)))=0$.

2) Show that $\lambda(f^{-1}((1,\infty)))=0$.

3) What can we conclude about $f$?

I have already done part a) of this problem, since we can choose $B=supp(f)$. Also, part b) 2) follows from part b)1) by monotony and $\sigma$-subadditivity since we have that $(1,\infty)=\bigcup_{n\in \mathbb{N}}[1+\frac{1}{n} , \infty)$. Also, what we can conclude about $f$, I believe, it's that from b)2) we would have that the set of elements such that the image belongs to $(1,\infty)$ has measure $0$, so we could apply part a) again and that condition will hold again, so $f$ can be seen almost everywhere again as a characteristic function in a borel set of finite measure.

If that is correct, my only trouble would be proving part b)1), and here I don't really know what to do, so any hint would be very appreciated. Thank you!

Answer 1

Hint : Let $A= f^{-1}([1+\alpha, \infty ))$. Then $\int_A f^{2n} \mathrm{d}\lambda \leq \int_\mathbb{R} f\mathrm{d}\lambda$.

But also $\int_Af^{2n} \mathrm{d}\lambda \geq (1+\alpha)^{2n} \lambda(A)$

Answer 2

a) First observe that, $$0\le f\le 1 \implies f(1-f)\ge 0$$

Then this yields $$\int_{ \mathbb{R} } f^2 d \lambda = \int_{ \mathbb{R} } f d \lambda \implies \int_{ \mathbb{R} } f(1-f)d\lambda = 0 \implies f(1-f) = 0~~~a.e$$

Therefore there exists $N$ a Borel set such that, $\lambda(N) = 0 $ and $$f(x)(1-f(x)) = 0~~~~\forall ~~x\in\Bbb R\setminus N$$ Then set $$A= \{x\in\Bbb R\setminus N: f(x) = 1\} $$ then we have $$A^c= \{x\in\Bbb R\setminus N: f(x) = 0\}\cup N $$

Then see that

• A is a Borel set since $f$ is measurable.
• if $x\in A$ we have $f(x) = 1$
• if $x\in A^c$ then we have $f(x) = 0 $ or $x\in N$ but $\lambda(N) = 0 $ Hence, $$ f(x)= \chi_A(x) ~\text{for all}~\in R\setminus N ~and ~~\lambda(N) = 0 \\\Longleftrightarrow f(x)= \chi_A(x) ~~\text{almost everywhere}$$ We also have, $$0\le \lambda(A) =\int_A d\lambda = \int_A 1 d\lambda+\int_{A^c} 0d\lambda+\int_N fd\lambda =\int_\Bbb R fd\lambda <\infty.$$

Since $f = 0~~a.e$ on $A^c$ $\lambda(N) = 0$ implies $\int_N fd\lambda = 0$.

Conclusion $f$ is the characteristic function of Borel set $$ f(x)= \chi_A(x) ~~\text{almost everywhere and}~~~\lambda(A) =\int_\Bbb R fd\lambda <\infty$$ since $f\in L^1,~~~with ~~~f\ge 0$

b) Setting $A_k= f^{-1}([1+\frac1k, \infty ))$ and $A= f^{-1}([1, \infty ))$ we have

$$A= \bigcup_{k=0}^{\infty} A_k$$ and and since $ f\ge 1\frac1k$ on $A_k$ we have, $$ (1+\frac1k)^{2n} \lambda(A_k) =\int_{A_k}(1+\frac1k)^{2n} d\lambda \le \int_{A_k} f^{2n} d\lambda \le \int_\Bbb R f^{2n} \mathrm{d} = \int_\mathbb{R} f\mathrm{d}\lambda.~~~\forall ~n\ge 1$$

that is $$ \lambda(A_k) \le(1+\frac1k)^{-2n}\int_\mathbb{R} f\mathrm{d}\lambda.~~~\forall ~n\ge 1$$ But since $f\in L^1,~~f\ge 0$, $$\int_\mathbb{R} f\mathrm{d}\lambda<\infty$$ Hence $$\lambda(A_k) \le\lim_{n\to\infty}(1+\frac1k)^{-2n}\int_\mathbb{R} f\mathrm{d}\lambda. = 0 $$

Hence $$\lambda(A_k)= 0~~~\forall ~~k>1~,\implies \lambda(A) =0 $$

Thus, (1) and (2) are solve.

for question 3., for all $n\ge 1, $

$$\int_{ \mathbb{R} } f^{2n} d \lambda = \int_{ \mathbb{R} } f d \lambda \implies \int_{ \mathbb{R} } f^{2n-1} (1- f )d \lambda = 0 .$$ Then , $$ 0=\int_{ \mathbb{R} } f^{2n-1} (1- f )d \lambda =\int_{ A } f^{2n-1} (1- f )d \lambda +\int_{ A^c } f^{2n-1} (1- f )d \lambda. $$

But since $\lambda(A)= 0$ and on A $1-f\le 0 $ we have, $$\int_{ A } f^{2n-1} (1- f )d \lambda = 0 \implies \text{for n=1}~~f(1-f) = 0~~a.e~~on ~~A $$ and also on $A$ we have $1-f\ge 0 $ along with $$\int_{ A } f^{2n-1} (1- f )d \lambda = 0$$ yield $$ 0=\int_{ \mathbb{R} } f^{2n-1} (1- f )d \lambda =\int_{ A } f^{2n-1} (1- f )d \lambda +\int_{ A^c } f^{2n-1} (1- f )d \lambda\implies \int_{ A^c } f^{2n-1} (1- f )d \lambda\ =0\\implies~~for~~ f(1-f)=0~~~a.e ~~~ ~~on~~~A^c$$

Conlsuion $f(1-f) = 0~~a.e~~on ~~A $ and $f(1-f) = 0~~a.e~~on ~~A^c $ means that $$f(1-f) = 0~~a.e~~~on~~\Bbb R.$$ Similar reasoning as in the first part a) shows that $f$ is the characteristic function of a Borel set with finite measure.