For a positive vector $\alpha\in\mathbb{R}^n$ ($n\geq 1$), denote by $\text{Dir}(\alpha)$ the Dirichlet distribution with parameter $\alpha$. In terms of weak convergence, is it true that, if $\sum\limits\_{i=1}^n\alpha\_i=1$, then $\lim\limits\_{\varepsilon\rightarrow 0^+}\text{Dir}(\varepsilon\alpha)\longrightarrow \sum\limits\_{i=1}^n \alpha\_i \delta\_{\lbrace e\_i\rbrace}$ (where $(e\_i)\_{1\leq i\leq n}$ is the canonical base of $\mathbb{R}^n$)?



Weak convergence of Dirichlet distributions to a "multi-Bernoulli" distribution
$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$For any $a=(a\_1,\dots,a\_n)\in(0,\infty)^n$ and any real $t\in(0,1/2)$, let $X=(X\_1,\dots,X\_n)$ have the Dirichlet distribution with parameter $ta$. Then $X\_1$ has the beta distribution with parameters $ta\_1$ and $tb\_1$, where $b\_1:=s-a\_1$ and
$$s:=a\_1+\dots+a\_n.$$


Let $t\downarrow0$. Then $\Ga(t)=\Ga(1+t)/t\sim1/t$ and hence  

$$P(X\_1>1-t)=\frac{\Ga(ts)}{\Ga(ta\_1)\Ga(tb\_1)}\,J
\sim\frac{ta\_1b\_1}s\,J,$$
where
$$J:=\int\_{1-t}^1 x^{ta\_1-1}(1-x)^{tb\_1-1}\,dx
\sim\int\_{1-t}^1 (1-x)^{tb\_1-1}\,dx=\frac{t^{tb\_1}}{tb\_1}\sim\frac1{tb\_1},$$
so that $P(X\_1>1-t)\to\dfrac{a\_1}s$. Similarly, for each $j\in[n]:=\{1,\dots,n\}$,
$$P(X\_j>1-t)\to\dfrac{a\_j}s.$$
Hence,
$$P(X\_j\le 1-t\ \forall j\in[n])\to1-\sum\_{j=1}^n\dfrac{a\_j}s=0.$$


So, for any continuous function $f\colon\R^n\to\R$,
$$Ef(X)=\sum\_{j=1}^n Ef(X)1(X\_j>1-t)+Ef(X)1(X\_j\le 1-t\ \forall j\in[n])
\to\sum\_{j=1}^n f(e\_j)\dfrac{a\_j}s+0,$$
where $e\_j$ is the $j$th standard basis vector of $\R^n$; here we used the implications $X\_j>1-t\iff1>X\_j>1-t\implies0<X\_i<t<1-t\ \forall i\in[n]\setminus\{j\}$.


Thus, the Dirichlet distribution with parameter $ta$ converges to $\sum\limits\_{j=1}^n \dfrac{a\_j}s \delta\_{\{e\_i\}}$ as $t\downarrow0$.
That is, your conjecture holds iff $s=1$.


