I am confused about a certain behaviour of reduced defect measures, and I would like to seek clarification on it. Below are the details.
Let $B$ denote the unit ball in $\mathbb{R}^n$. Suppose that $\{f_m\}_{m\in\mathbb{N}}\subset L^2(B)$ is a uniformly bounded sequence (in the $L^2$-norm) such that \begin{equation} \left\{\begin{array}{l} f_m\rightarrow f\quad\text{a.e. in }B\\ f_m\rightharpoonup f\quad\text{in }L^2(B) \end{array}\right. \end{equation} as $m\rightarrow\infty$. We cannot conclude that $f_m\rightarrow f$ in $L^1(B)$ as $m\rightarrow\infty$. To examine this failure of strong convergence DiPerna and Majda introduce the so called reduced defect measure: \begin{equation} \theta(E)\equiv\limsup_{m\rightarrow\infty}\int_{E}|f_m-f|^2\ \mathrm{d}x\quad E\subset B,\ E\text{ is Borel}, \end{equation}which I suppose when defined as: \begin{equation} \theta(E)\equiv\inf\left\{\limsup_{m\rightarrow\infty}\int_G|f_m-f|^2\ \mathrm{d}x\ \Big|\ E\subset G, \ G\text{ is Borel}\right\}\quad (\theta\subset B) \end{equation}gives a measure on $B$. However, $\theta$ fails the countable sub-additivity property in general. I suppose this is because we don't have a mechanism for swapping the limits involved. Nonetheless, $\theta$ is finitely sub-additive.
It seems to me that $\theta$ is absolutely continuous with respect to $\mathcal{L}^n$, because if $E\subset B$ is such that $|E|=0$, then (as $\mathcal{L}^n$ is Borel regular) there is a Borel set $G$ such that $E\subset G$ and $|G|=0$. Consequently, \begin{equation} \int_{G}|f_m-f|^2\ \mathrm{d}x=0\quad\text{for all } m\in\mathbb{N}, \end{equation}and therefore, \begin{equation} \theta(E)\leq \limsup_{m\rightarrow\infty}\int_{G}|f_m-f|^2\ \mathrm{d}x=0. \end{equation}
There is a known theorem in measure theory that says if $\mu$ is a finite measure on a set $X$ and $\{E_{\alpha}\}_{\alpha\in I}$ is a collection of pairwise disjoint measureable sets in $X$, then there are at most countably many $E_{\alpha}$ such that $\mu(E_{\alpha})>0$. The proof of this theorem does not call upon the countable sub-additivity property of $\mu$, only finite sub-additivity.
As $\theta$ is finitely sub-additive and $\theta(B)<\infty$, the above theorem is applicable for $\theta$. Thus, if $\theta(B)>0$, and we write \begin{equation} B=\bigcup_{y\in B}\{y\}, \end{equation}we have $\{\{y\}\}_{y\in B}$ is a pairwise disjoint collection of measurable sets in $B$ and only countably many are such that $\theta(\{y\})>0$. In particular, there is at least one $\{y\}$ such that $\theta(\{y\})>0$.
However, $|\{y\}|=0$, which in turn should imply that $\theta(\{y\})=0$ - an apparent contradiction.
I would like help figuring out what went wrong with my argument.
Your issue is in the line: ... and only countably many are such that $\theta(\{y\})>0$. In particular, there is at least one $\{y\}$ such that $\theta(\{y\})>0$.
The same issue comes up if you use the Lebesgue measure instead of $\theta$. We can write $B=\bigcup_{y\in B}\{y\}$, and we have $0 < \mathcal{L}^N(B) < \infty$. Do you see the issue with deducing from this that there is at least one point with positive Lebesgue measure? Of course this isn't possible since we can easily show and $\mathcal{L}^N(\{y\})=0$ for all $y\in B$.
The theorem you reference only says that there are at most countably many positive measure sets in your collection. However, it does not guarantee the existence of even one such set.