= Solution
<Weak convergence> $u_j\rightharpoonup u$ means $F(u_j)\to F(u)$ for every $F\in X^*$. <weak-star topology>[Weak-star convergence] $T_j\overset*\rightharpoonup T$ means $T_j(x)\to T(x)$ for every $x\in X$, viewing $X$ as the predual. The <Sequential Banach-Alaoglu theorem for a separable predual> says that every norm-bounded <sequence> in $X^*$ has a <weak-star topology>[weak-star convergent] <subsequence>. Separability is what turns the <bounded> <weak-star topology>[weak-star] <compactness> statement into this sequential form.
For the <weak-star limit of periodic rescalings>, apply it with $X=L^1(\mathbb R)$, which is separable, and $X^*=L^\infty(\mathbb R)$. The rescaled <functions> have uniform <essential supremum>[essential-supremum] bound $\|\theta\|_\infty$. Let $q=\theta-\bar\theta$ and $F(t)=\int_0^tq(s)\,ds$. Since $q$ is one-periodic with zero mean, $F$ is a <bounded> one-periodic absolutely <continuous> primitive. Thus for every finite <interval>,
$$
\int_a^b[\theta(jx)-\bar\theta],dx=\frac{F(jb)-F(ja)}j\longrightarrow0.
$$
Every <weak-star topology>[weak-star subsequential limit] $h$ therefore satisfies $\int_a^b(h-\bar\theta)=0$ for all $a<b$. To apply the supplied uniqueness fact without assuming global integrability, fix $R>0$ and <set> $v=(h-\bar\theta)\mathbf1_{[-R,R]}$. This is in $L^1(\mathbb R)$, and its <integral> over any <interval> is zero, since intersecting with $[-R,R]$ again gives an <interval>. Hence $v=0$ <almost everywhere>. Taking integer $R\to\infty$ yields $h=\bar\theta$ <almost everywhere> on the whole line.
Finally, if the full <sequence> did not converge <weak-star topology>[weak-star] to this constant, some $g\in L^1$ and some <subsequence> would have its pairing with $\theta(jx)-\bar\theta$ <bounded> away from zero. Sequential <compactness> gives a further <weak-star topology>[weak-star convergent] <subsequence>, whose <limit> must be the constant just identified, a contradiction. Therefore
$$
\boxed{\theta(j\cdot)\overset*\rightharpoonup\bar\theta\quad\text{in }L^\infty(\mathbb R).}
$$
This is averaging against integrable <test functions>, not a claim of <pointwise> or strong <norm convergence>.
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