Can a "continuous" convex combination not be element of the convex hull?

As others have said the two sets are the same. The fact that $\Gamma\subset \Lambda$ essentially follows from the fact that a convex combination $\sum_1^k w_i a_i$ is equal to $\int a(x)w(x)\, dx$ with $w=\sum_1^k w_i\delta_{a_i}$ (Dirac deltas concentrated at $a_i$).

The opposite inclusion follows from Jensen's inequality. Consider the function (this is called characteristic (or indicator) function in convex analysis) $$I_\Gamma(x)=\begin{cases} 0 & x\in \Gamma \\ +\infty & x\notin \Gamma\end{cases}$$ This function is convex and $\Gamma=\{x\ :\ I_\Gamma(x)=0\}$. Now let $\lambda = \int a(x)w(x)\, dx\in\Lambda$. By Jensen's inequality $$ I_\Gamma(\lambda)\le \int I_\Gamma(a(x))w(x)\, dx=0, $$ so $I_\Gamma(\lambda)=0$, which means that $\lambda \in \Gamma$.

One solution is to consider the following characterization of convex sets:

A closed set is convex if and only if it is the intersection of closed half-spaces.

So, all half-spaces are of the form $x\cdot v \geq c$, so if we want to prove this in the case where $\Gamma$ is closed, all that we need is the following: If $a(x)\cdot v\leq c$ everywhere, then $\int_{\Omega}w(x)a(x)\cdot v\,dx\leq c$ for any distribution $w$. However, this is trivial by monotonicity: That integrand cannot exceed $w(x)c$ so the integral cannot exceed $\int_{\Omega}w(x)c\,dx = c$, as desired. This tells us that $\Lambda$ is a subset of $\text{cl}(\Gamma)$.

To work out the boundary, you just note that for any point $p$ on the boundary, there is at least one plane through it not intersecting the interior. Moreover, if $w(x)$ assigns positive weight a portion of $a(x)$ not on this plane, then the integral will not be on this plane (since it will be strictly below it). Otherwise, $w(x)$ only assigns weight on that plane, in which case we're dealing with the two dimensional analog of the problem (since the intersection of that plane and $\Omega$ and $\Gamma$ acts as expected). So, we can work out an induction proof on dimension that will make sure the boundary works out.