Intuition of weak star convergence. Given $\Omega=(0,1)$, consider the following sequence
$$
v_j(x)\colon=\begin{cases}
\;a &\text{if }jx-\lfloor jx \rfloor\le\theta\\
\;b &\text{otherwise}
\end{cases}
$$
where $a,b\in\mathbb{R}$ and $\theta\in(0,1)$. Then the $L^\infty-\text{weak}^\ast$ limit of $v_j$ is $\theta a+(1-\theta)b$. Is there any intuitive explanation of why this is the weak$^\ast$ limit of $v_j(x)$?
 A: You can think of the weak* convergence as a convergence locally on average. 
Observe that the $v_j$'s oscillate assuming the value $a$ on intervals of length $\theta/j$ and value $b$ on intervals of length $(1-\theta)/j$, alternating between the two, i.e.
$$v_j(x)=\begin{cases} a& \mathrm{if }\; x\in \left[\frac{k}{j}, \frac{k+\theta}{j}\right]\\
b& \mathrm{if}\;x\in \left(\frac{k+\theta}{j}, \frac{k+1}{j}\right) 
\end{cases}.$$
By definition to compute the weak* limit you have to test the $v_j$'s against any $L^1$ function, so let's start with the characteristic of an interval $(s,t)$. If we take $(s,t)=(\tfrac{k}{j},\tfrac{k+1}{j})$ with $k\in\mathbb{N}$ it is easy to see that
$$\int\limits_0^1 \chi_{(s,t)}(x)v_j(x)dx=\int\limits_s^t v_j(x)dx=\frac1j \left(\theta a+(1-\theta)b\right).$$
If now we take a generic interval $(s,t)\subset (0,1)$, the number $N(j,s,t)$ of intervals of the form $(\tfrac{k}{j},\tfrac{k+1}{j})$ contained in $(s,t)$ is roughly $(t-s)/(1/j)=j(t-s)$ (with an error of at most $2$), so we have
\begin{align}
\int\limits_0^1\chi_{(s,t)}(x)v_j(x)dx &=\int\limits_s^t v_j(x) dx\\
&= N(j,s,t) \frac1j \big(\theta a+(1-\theta)b\big)+o(1)_{(j\to\infty)}\\
&=(t-s)\big(\theta a+(1-\theta)b\big)+o(1)_{(j\to\infty)}
\end{align}
where $o(1)_{(j\to\infty)}$ is a term that goes to $0$ as $j\to\infty$ and is due to the two spare intervals near $s$ and $t$ that we are not counting among $N(j,s,t)$, and to the small $\pm 2$ error in $N(j,s,t)$ itself. In this way we obtained that for any characteristic test function $\chi_{(s,t)}$ it holds
$$\lim_{j\to\infty}\int\limits_0^1 \chi_{(s,t)}(x)v_j(x)dx=(t-s)\big(\theta a+(1-\theta)b\big)=\int\limits_0^1 \chi_{(s,t)}(x)\big(\theta a+(1-\theta)b\big)dx$$
so that the constant function $\big(\theta a+(1-\theta)b\big)$ is a good candidate as the weak* limit. We just have to check that this holds not only for characteristic functions of intervals but for any $L^1$ function. To see this you can use that characteristics of intervals are dense in $L^1$ and conclude by approximation.
In general to compute a weak* limit a good way is to test against characteristic functions of quite simple sets, for instance intervals on $\mathbb R$, and this will give you the average of the $v_j$'s on that set, thus telling you what is locally the weak* limit.
