I have a question regarding the following partition of a hypercube $H_{R}(x)$ centered at $x$ with sides of length $R$ in $\mathbb{R}^{n}$:
Consider this hypercube $O = H_{R}(x) = \prod_{i=1}^{n}(\alpha_{i},\beta_{i}) = (\alpha_{1},\beta_{1}) \times \prod_{i=2}^{n}(\alpha_{i}, \beta_{i})$ and the following partition of it:
We divide $(\alpha_{1}, \beta_{1})$ into intervals of length $(\beta_{1}-\alpha_{1})2^{-m}$ where $m \in \mathbb{N}$. Each of these intervals is then subdivided into two parts of equal length $(\beta_{1}-\alpha_{1})2^{-(m+1)}$. The union of the first(resp. the second) subintervals is denoted by $I_{m}$(resp. $J_{m}$). We denote by $$O_{1}^{m} = I_{m} \times \prod_{i=2}^{n}(\alpha_{i}, \beta_{i})$$ $$O_{2}^{m} = J_{m} \times \prod_{i=2}^{n}(\alpha_{i}, \beta_{i})$$ then $\text{meas } O_{1}^{m} = \text{meas }O_{2}^{m} = \frac{1}{2}\text{meas } O$
Can anyone see how it would follow that, if considering the characteristic functions $\chi_{O_{1}^{m}}$ and $\chi_{O_{2}^{m}}$ on the sets $O_{1}^{m}$ and $O_{2}^{m}$ respectively, that we get the result that $\chi_{O_{1}^{m}} \rightharpoonup^{*} \frac{1}{2}$ and $\chi_{O_{2}^{m}} \rightharpoonup^{*} \frac{1}{2}$ in $L^{\infty}(O)$.
Thanks for any assistance, let me know if anything is unclear.
You have a bounded sequence $f_m$ in $L^\infty$, specifically $\|f_m\|_{L^\infty}\le 1$ for all $m$. For such a sequence, to verify that $f_m\xrightarrow{w^*} f $ it suffices to check that $$\langle g,f_m\rangle \to \langle g,f\rangle \tag{1}$$ for all $g$ in some set $A\subset L^1$ with norm-dense linear span. Indeed, once we have (1) for $g\in A$, we have it for linear combinations of elements of $A$. And then, given any $h\in L^1$, we can find $g$ in the span of $A$ such that $\|h-g\|_{L^1}\le \epsilon$. It follows that $$ | \langle h,f_m\rangle - \langle h,f\rangle | \le 2\epsilon + | \langle g,f_m\rangle - \langle g,f\rangle |$$ and since $\epsilon$ is arbitrarily small, $| \langle h,f_m\rangle - \langle h,f\rangle | \to 0$.
It remains to pick a convenient set $A$ with norm-dense span in $L^1$. I suggest the set of characteristic functions of dyadic rectangles. With this choice, you'll see that for every $g\in A$, $$\int gf_m = \frac12 \int g$$ for all $m$ large enough. This shows (1),