Defect measure associated to a sequence of exponentials

Question

In my road to understand microlocal defect measures, at the beginning of Gerard's article Microlocal defect measures, there is an statement about (an example of) defect measures where I am struggling.

The context is the following. Let $\Omega$ be an open set in $\mathbb{R}^d$ and let $(u_n)_n$ be a bounded sequence in $L_{loc}^2(\Omega)$ which converges to $u\in L_{loc}^2(\Omega)$ in the sense of the distributions. By identifying $L^1$ with a subspace of the space of Radon measures, a weak-$*$ argument allows us to deduce that $$|u_n-u|^2\rightharpoonup \nu$$ where $\nu$ is a Radon measure, called the defect measure associated to $(u_n)_n$.

To exemplify that the defect measures does not capture some important properties, the sequence $u_n(x)=\exp(inx\cdot \xi)$, for some $\xi\in \mathbb{R}^d\setminus\{0\}$, is examined. This sequence is bounded in $L_{loc}^2(\mathbb{R}^d)$, $u_n\rightharpoonup 0$ in the sense of the distributions and does not converge strongly in $L_{loc}^2$. The defect measure $\nu$ associated to this sequence is the Lebesgue measure, for any $\xi\in \mathbb{R}^d$.

My question is: how do I find that, indeed $\nu=\lambda$? I've been thinking about it, but I did not be able to unravel the statement. Any hint is well received!

Answer 1

Assuming that $\lambda$ in your notation stands for $d$-dimensional Lebesgue measure, we can observe simply that the sequence of interest to you is $$ |u_n(x)-u(x)|^2=|\exp(inx\cdot\xi)|^2=1, $$ which then is the Lebesgue measure. If it is not clear why this is, simply write the definition of $|u_n|^2\rightharpoonup \lambda $.