Let $f:X\to Y$ be a Borel Map from the measure space $(X,\mathcal{B},\mu)$ to the measurable space $(Y,\mathcal{M})$. The pushforward of $\mu$ denoted by $f_\#\mu$ is defined as $f_\#\mu(E)=\mu(f^{-1}(E))$ for $E\in\mathcal{M}$.
Let $C([0,1];\mathbb{R}^d)$ be the space of all continuous functions defined on $[0,1]$ with the supremum norm. Let $t\in[0,1]$ and let $e_t$ denote the evaluation map $e_t:C([0,1];\mathbb{R}^d)\to \mathbb{R^d}$ given by $e_t(f)=f(t)$.
Question: I recently read in a paper, pp14, Exercise 14, that a measure $\sigma$ on $C([0,1];\mathbb{R}^d)$ is a Dirac measure iff the pushforward measures $e_t{_\#}\sigma$ are Dirac measures on $\mathbb{R^d}$ for all $t\in\mathbb{Q}\cap[0,1]$. One way is trivial. The other way I cannot prove.
The obvious way to proceed, I thought, was to construct a function
$f:[0,1]\to \mathbb{R^d}$ by $f(t)=x_t$ for $t\in\mathbb{Q}\cap[0,1]$ where $e_t{_\#}\sigma=\delta_{x_t}$ for some $x_t\in\mathbb{R^d}$. Then, I thought of defining the function for $t\notin\mathbb{Q}\cap[0,1]$ by choosing a rational sequence $t_n$ converging to $t$ and looking at the corresponding sequence $x_{t_n}$. But without any form of continuity available for $f$ on $\mathbb{Q}\cap[0,1]$ I cannot conclude that this sequence $x_{t_n}$ is Cauchy, without which I cannot define $f$ on $[0,1]$.
Could someone please help!
Thank you.