Let $S$ be a closed convex set (of a Banach space, not necessarily of finite dimension), $(\Omega, \mathcal{B}(\Omega), \mu)$ be a probability space, and $\phi : \Omega \to S$ be an integrable function (or a random variable of finite expectation if you prefer).
Surely, if $\mu$ is countable, we can write it, $$ \mu = \sum_{n \in \mathbb{N}} p_n \delta_{x_n}, $$ for some sequence $(x_n)_{n \in \mathbb{N}}$ of elements in $\Omega$ and nonnegative weights $(p_n)_{n \in \mathbb{N}}$, summing to $1$. Then, clearly, for all $n \in \mathbb{N}$, by convexity of $S$ $$ \frac{\sum_{k \leq n} p_k \phi(x_k)}{\sum_{k \leq n} p_k} \in S, $$ then by closure of $S$, the limit belongs to $S$.
My question is, does this statement hold when $\mu$ is not supposed countable? Namely, is the following always true? $$ \int_\Omega \phi \mathrm{d}\mu \in S $$
Approximate $\phi$ by simple functions. In particular, for each $\epsilon>0$ you can use your argument above to show $\int \phi d\mu\in S_\epsilon$ where $S_\epsilon$ is the set of points at distance $\le\epsilon$ to $S$. The sets $S_\epsilon$ are closed and convex, and $S=\bigcap_{\epsilon>0}S_\epsilon$.
Added in an edit. Or you could use the Hahn-Banach theorem. If $\mu$ is not in $S$ there is a continuous linear function $\lambda$ that separates them, and from that a contradiction is easy to exhibit: $\langle \lambda,\phi \rangle\ge a$ with probability $1$ but $\langle \lambda,\mu\rangle\le b$ for some real $b<a$.