continuity - Continuous and almost everywhere continuously differentiable with bounded gradient implies Lipschitz?

Let $f:\mathbb{R}^n\to \mathbb{R}$ be continuous and bounded. Let the gradient of $f$ be continuous almost everywhere and let it be uniformly bounded wherever it is defined, i.e. $\|\nabla f(x)\|\le M$ for any $x\in\mathbb{R}^n$, where $M$ is a positive constant. Can we say that $f$ is globally Lipschitz continuous?

If so, where can we find a formal proof?

If not, can we provide a counterexample? (I.e., a function $f:\mathbb{R}^n\to \mathbb{R}$ that is continuous and bounded, has uniformly bounded gradient but is not Lipschitz.)

Remark.
This question has been answered (positively) for $n=1$ in If $f$ is continuous and piecewise $C^1$ and $f'$ is bounded a.e., is $f$ Lipschitz?.

Remark.
This question has been answered (positively) for $f$ continuously differentiable in A continuously differentiable map is locally Lipschitz. However, from the answer given there, I seem to understand that continuity of the derivatives can be relaxed to boundedness of the derivatives. If this is actually the case, then my question is also answered (positively) in that post. But judging from the following remark, I may be missing something.

Remark
This paper http://www.sciencedirect.com/science/article/pii/S0024379512002741
seems to prove that there DO exist counterexamples (i.e. functions that are continuous, have uniformly bounded derivatives but are NOT Lipschitz). However, I must confess I do not understand their proof.

Answer

As stated, the answer to your question is no. The Cantor function is a common counterexample when the derivative is required to exist only almost everywhere. It is continuous, has zero derivative a.e., in particular in an open set of full measure, but it is not Lipschitz continuous, nor absolutely continuous. You can make an analogous example in the cube $\mathbb{R}^n$ considering the Cantor function of the first variable.

If instead you take a function $f\in L^1_{loc}(\mathbb{R}^n)$ and require its gradient to exist in the distributional sense and to be represented by a function in $L^\infty$, then you are by definition stating that $f$ belongs to the Sobolev space $W^{1,\infty}(\mathbb{R}^n)$, which can be proven to coincide with the space of Lipschitz functions, see for example this question. This still holds if you replace $\mathbb{R}^n$ with a convex set $\Omega$ (there are some weaker conditions on $\Omega$ under which it works, but it doesn't work with arbitrary domains). Note that you don't need to assume a priori that $f$ is continuous, nor that the gradient is continuous almost everywhere.