I have read some papers that give the definition of one-wayness of PKE schemes.
Let $\Pi = (G,E,D)$ be a PKE scheme, and the security game of OW-CPA is defined as follow:
$$\mathrm{Adv}_{\Pi,\mathcal{A}}^{\text{ow-cpa}}(k) = \Pr\left[ m = m' \left\vert \begin{gathered} (pk,sk) \gets G(1^k) \\ m \gets \mathcal{M}_{k} \\ c \gets E_{pk}(m) \\ m' \gets \mathcal{A}(pk,c) \\ \end{gathered}\right.\right]$$
We say $\Pi$ is secure in the sense of OW-CPA if $\mathrm{Adv}_{\Pi,\mathcal{A}}^{\text{ow-cpa}}(\cdot)$ is nelgibile.
My question is that how does the challenger choose $m$ over $\mathcal{M}_{k}$ randomly? Is $m$ an element picked uniformly from $M_{k}$?
If $\mathcal{M}_{k} = \mathcal{M} = \{0,1\}^*$, $m$ cannot be picked uniformly from $M_{k}$.
If $|\mathcal{M}_{k}| = \mathrm{poly}(k)$, then $\Pi$ cannot be secure in the sense of OW-CPA if $\mathcal{D}_{k}$ is uniform.
Let $F$ be the sample circuit, then $$\mathrm{Adv}_{\Pi,\mathcal{A}}^{\text{ow-cpa}}(k) = \Pr\left[ m = m' \left\vert \begin{gathered} (pk,sk) \gets G(1^k) \\ m \gets F(\mathcal{M}_{k}) \\ c \gets E_{pk}(m) \\ m' \gets \mathcal{A}(pk,c) \\ \end{gathered}\right.\right]$$
Assume that the distribution of message is $\mathcal{D}_{k}$, thus $m$ follows $\mathcal{D}_{k}$. We cannot define that $\Pi$ is OW-CPA secure if for every distribution $\mathcal{D}_{k}$, $\mathrm{Adv}_{\Pi,\mathcal{A}}^{\text{ow-cpa}}(\cdot)$ is negligible, because it is not OW-CPA secure if $\mathcal{D}_{k}$ is a one-point distribution.
Perhaps, we may define the PKE scheme as $\Pi = (G,E,D,F)$ where the sample circuit $F$ is given at first. But the notion of IND-CPA or SS-CPA does not need $F$.
[FO99] Secure Integration of Asymmetric and Symmetric Encryption Schemes