I deem a generalized framework formalizing the concepts at work is apt here. For more details, refer to $\rm [I].$
Let $(\Omega, \boldsymbol{\mathfrak A}, \Pr)$ be a probability space. Consider the sequence of probability spaces $\langle (\mathcal X_i, \boldsymbol{\mathfrak A}_i, \mathbf P_i)\rangle_{i=1}^\infty,$ where $(\mathcal X_i, \Vert \cdot \Vert_i)$ is a normed linear space.
Consider a sequence of rvs $\langle X_i\rangle_{i=1}^\infty$ and a sequence of real numbers $\langle r_i\rangle_{i=1}^\infty.$ Then $X_n = o_P(r_n)\iff \Pr[\Vert X_n \Vert_n\leq c|r_n|] = 1, ~\forall c >0.$
Now consider a sequence of measurable functions $f_n:\mathcal X_n\to \mathcal R ,~\mathcal R$ being a normed linear space with Borel $\sigma$-field. Define $T_n := f_n(X_n)$ and $T: \Omega \to \mathcal R. $ Then $T_n$ converges in probability to $T$ if and only if $\Vert T_n - T\Vert = o_P(1).$
Now consider a parametric family of distributions $\{\mathbf P_\theta\mid \theta\in \Theta \}$ on a sequence space $\mathcal X^\infty.$ Define a measurable function $g: \Theta\to \mathcal G, ~\mathcal G$ being a metric space with Borel $\sigma$-field. Take $\mathcal X_n = \mathcal X^n$ and take measurable functions $T_n: \mathcal X_n\to \mathcal G.$ Then $T_n$ is consistent for $g(\theta)$ if for each $\theta, ~T_n\overset{\mathbf P}{\to} g(\theta).$
The simplest and most common instance is taking $\Omega = \mathbb R^\infty, ~\mathcal X_n = \mathbb R^n.$
Observe how the underlying probability space is at work here based on the implications of the characterization of the convergence in probability above.
(Also, as a footnote, one can see how in probability can be generalized: Take $S\subseteq \prod_{i=1}^\infty \mathcal X_i.$ Then $S$ occurs in probability (denoted by $\mathcal P(S)$) if, for each $i,$ there exists $S_i(\varepsilon)\in\boldsymbol{\mathfrak A}_i$ such that $\prod_{i=1}^\infty S_i(\varepsilon)\subseteq S$ and for each $\varepsilon > 0,~\mathbf P_i(S_i(\varepsilon))\geq 1-\varepsilon. $ To see how powerful it is, consider $f_n:\mathcal X_n \to \mathbb R$ and, as above, take $T_n = f_n(X_n).$ Now, define $S:= \left\{\langle x_i\rangle_{i=1}^\infty\mid\lim_{n\to\infty} f_n(x_n) = 0\right\}.$ Then $T_n = o_{\mathbf P}(1)\iff \mathcal P(S).$)
Reference:
$\rm [I]$ Theory of Statistics, Mark J. Schervish, Springer-Verlag, $1995,$ sec. $7.1.2,$ pp. $395-398.$