0
$\begingroup$

In a book concerning calculus of variations written by Giusti, I read the following
" Let $Q$ be a cube. For every $\lambda\in [0,1]$ there exists a sequence $\chi_h$ of characteristic functions of measurable sets $E_h\subset Q$ such that $$\chi_{h}\to \lambda \chi_Q$$ in the weak-star topology of $L^{\infty}$."

I don't know why the sequence exists. I have known from Banach–Alaoglu–Bourbaki Theorem that there exists a sequence $f_h$ with $\lVert f_h\rVert_{L^\infty}\leq 1$ such that $$f_h\to \lambda\chi_Q$$ in the weak-star topology. But how can we rewrite $f_h$ as a characteristic function?

$\endgroup$
2
  • $\begingroup$ What is the setting? $\mathbb{R}^n$ with Lebesgue measure, or something more exotic? $\endgroup$ Commented Sep 20 at 12:44
  • $\begingroup$ $\mathbb{R}^n$ with Lebesgue measure. $\endgroup$ Commented Sep 20 at 13:02

1 Answer 1

1
$\begingroup$

I suppose the underlying measure space is the Lebesgue measure on $\mathbb R^d$.

Here is a sketch of the proof using the Lyapunov convexity theorem.

Let $(v_n)$ be a dense set of $L^1(\mathbb R^d)$. Define $v_0:=\chi_Q$. Then each $v_i$ defines a non-atomic measure via $A \mapsto \int_A v_i\ dx$.

Fix $n$. Then by Lyapunov theorem, there is a set $E_n$ such that $$ \int_{E_n} v_i \ dx =\lambda \int_Q v_i \ dx \quad \forall i =0 \dots n. $$ This creates the sequence $\chi_n := \chi_{E_n}$ of characteristic functions.

Now take $v\in L^1(\mathbb R^d)$ and $\epsilon>0$. Then there is $v_i$ with $\|v-v_i\|_{L^1} \le \epsilon$. Take $n>i$. Then $$ \int_{\mathbb R^d} (\chi_{E_n} - \lambda \chi_Q) v \ dx= \int_{\mathbb R^d} (\chi_{E_n} - \lambda \chi_Q) (v-v_i) \ dx + \int_{\mathbb R^d} (\chi_{E_n} - \lambda \chi_Q) v_i \ dx. $$ By construction of $E_n$ and $n>i$, the second addend is zero. The first is less than $\epsilon$: $$ \left| \int_{\mathbb R^d} (\chi_{E_n} - \lambda \chi_Q) (v-v_i) \ dx \right| \le \|\chi_{E_n} - \lambda \chi_Q\|_{L^\infty} \|v-v_i\|_{L^1} \le 1 \cdot\epsilon. $$ This proves $\chi_{E_n} \rightharpoonup^* \lambda \chi_Q$ in $L^\infty$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.