This doesn't answer the original question; instead, it argues that the prohibition against using *any* derivatives of the 'candidate' curvature form is not really necessary. The first step is to show that, if a continuous $\frak{g}$-valued $2$-form $\Phi$ is the curvature of a continuous $\frak{g}$-valued $1$-form $\alpha$ (which must necessarily have a continuous exterior derivative), then $\Phi$ itself must have a continuous exterior derivative and the Bianchi identity must hold. (Note that this is weaker than claiming that $\Phi$ is differentiable, which might well not be true. It's easy for a differential form to be only $C^0$ but still have a continuous exterior derivative.)

I'm going to concentrate on the 'local' problem, i.e., I'm going to assume that the bundle is trivial. Thus, suppose that you have a continuous $2$-form $\Phi$ on a manifold $M$ that takes values in a Lie algebra $\frak{g}$. We want to know whether there is a $\frak{g}$-valued $1$-form $\alpha$ that is continuous (and has a continuous exterior derivative $d\alpha$) such that $\Phi = d\alpha + \frac12[\alpha,\alpha]$.

Let $\mathcal{C}_p(M)$ denote the vector space of smooth singular $p$-chains in $M$ with real coefficients, and let $\partial:\mathcal{C}_p(M)\to \mathcal{C}_{p-1}(M)$ denote the boundary operator. If the desired $\alpha$ exists, then, for every $\sigma\in\mathcal{C}_3(M)$, one has the identity
$$
\int_{\partial\sigma}\Phi = \int_{\partial\sigma} d\alpha + \frac12[\alpha,\alpha] = \int_{\partial\sigma}\frac12[\alpha,\alpha]= \int_{\sigma} [d\alpha,\alpha]=\int_{\sigma} [\Phi,\alpha].
$$
Note that I did *not* differentiate $\Phi$. This calculation shows that, if $\alpha$ exists, then $\Phi$ has to have a continuous exterior derivative, in the sense that there has to be a continuous $3$-form $\Psi$ such that
$$
\int_{\partial\sigma}\Phi = \int_\sigma \Psi
$$
for all $\sigma\in\mathcal{C}_3(M)$. Thus, a necessary condition that $\Phi$ be the curvature of a continuous connection form $\alpha$ having a continuous exterior derivative is that $d\Phi$ must exist and be continuous.

Moreover, this calculation shows that, if there is a continuous solution $\alpha$, it must satisfy the linear algebraic equation $[\Phi,\alpha] = d\Phi$. In particular, $d\Phi$ must lie in the image of the operator $W_\Phi: {\frak{g}}\otimes\Omega^1(M)\to
{\frak{g}}\otimes\Omega^3(M)$ defined by $W_\Phi(\alpha) = [\Phi,\alpha]$. This restriction frequently gives us algebraic equations that $(\Phi,d\Phi)$ must satisfy.

When ${\frak{g}} = {\frak{su}}(2)$ and the dimension of $M$ is $4$, the map $W_\Phi$ is an isomorphism for 'generic' $\Phi$ (where 'genericity' is a pointwise condition). In such a case, there will exist a unique $\alpha$ satisfying $[\Phi,\alpha] = d\Phi$ (and it will necessarily be continuous). If this $\alpha$ does not have a continuous exterior derivative or, if does have a continuous exterior derivative but doesn't satisfy $d\alpha + \frac12[\alpha,\alpha] = \Phi$, then there is no solution. Thus, in this `generic' case, one sees that the condition to be a curvature is that certain combinations of the first and second derivatives of $\Phi$ exist and provide a solution to the problem, just by differentiation.

When ${\frak{g}} = {\frak{su}}(2)$ and the dimension of $M$ is greater than $4$, the map $W_\Phi$ is an injection for 'generic' $\Phi$ (where 'genericity' is a pointwise condition), but it is not a surjection, so the condition $[\Phi,\alpha] = d\Phi$ already places first-order conditions on $\Phi$ and uniquely determines the only possible candidate $\alpha$, when it exists at all.

Other Lie algebras have to be treated on a case-by-case basis, to some extent, though there is probably something similar to the above story for all semi-simple Lie algebras.

areallowed to use if we can't differentiate? There is no point-wise condition on the curvature of a connection, since you can achieve any given Lie-algebra-valued $2$-form as a curvature at a point. If you take the simplest case, where the Lie algebra is $1$-dimensional, then you are asking us to tell you whether a given $2$-form is closed without allowing us to apply the exterior derivative. Can you tell us what you had in mind as a criterion in this simplest case? Maybe knowing that its integral over every boundary of a $3$-dimensional submanifold is zero? $\endgroup$ – Robert Bryant Dec 2 '11 at 13:22whywe can't use the Bianchi identity. Usually, you don't throw out useful tools unless you have some reason to believe that they can't be brought to bear (for example, you might be dealing with $2$-forms that are only continuous, a priori). However, Mircea seems not to be able/willing to divulge the motivation for solving this problem. $\endgroup$ – Robert Bryant Dec 2 '11 at 19:13