Intereting Posts

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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.

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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.

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If ${e_t}_\# \sigma = \delta_{x_t}$, then for each $t \in \mathbb{Q}$, the set $A_t := e_t^{-1}(\{x_t\}) = \{ f \in C([0,1], \mathbb{R}^d) : f(t) = x_t\}$ has $\sigma(A_t) = 1$. Therefore $A := \bigcap_{t \in \mathbb{Q}} A_t$ also has $\sigma(A)=1$; in particular $A$ is not empty. But if $f,g \in A$, we have $f(t) = x_t = g(t)$ for all $t \in \mathbb{Q}$, so by continuity $f=g$. Thus $A$ contains exactly one function $f$, and we have $\sigma = \delta_f$.

One has to note at some point that $\sigma$ is indeed a probability measure. Then, as an “alternative” to Nate’s solution, you can employ the language of stochastic processes: Call $X$ the canonical $\mathbb{R}^d$-valued process on the probability space $(C([0,1];\mathbb{R}^d),\mathcal{B},\sigma =: P)$ with distribution $\sigma$. Then $X$ has continuous paths and $X_t = x_t$ for all $t \in \mathbb{Q} \cap [0,1]$ $P$-a. s. But this means that $X$ is deterministic, so $\sigma$ must be a Dirac measure.

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