The answer posted by Tom, as written is actually not true. A function in $H^1$ will not in general be differential almost everywhere; it depends on the dimension. In one dimension however it is indeed true that $H^1$ functions are differentiable almost everywhere (they are in fact absolutely continuous). There are two ways of seeing it is not in $H^1$. The simple answer is that if you differentiate the characteristic function of say $[0,\infty)$ then you will get the dirac measure. However let me just answer your question first:
Answer 1: Take any smooth compactly supported $\phi:\mathbb{R} \to \mathbb{R}$. By definition of weak derivative we have $\int \phi g^{\prime} dx = - \int \phi^{\prime} g dx$ where I've set $g=1_{[0,\infty)}$. This would have to be true for all such $\phi$ if the weak derivative existed. Now take $\phi^{\epsilon}$ to be supported in a neighborhood $(-\epsilon,\epsilon)$ of $0$. We are making the crucial assumption that $g^{\prime}$ is an integrable and hence it follows that $\int \phi^{\epsilon} g^{\prime} \to 0$ as $\epsilon \to 0$. However, $\phi^{\epsilon}$ is smooth and so $\int \partial_x\phi^{\epsilon}(x)g(x)dx = \phi^{\epsilon}(0)$ since $\phi$ was assumed to have compact support in $(-\epsilon,\epsilon)$. Now just fix $\phi^{\epsilon}(0)=1$ and we have that $\phi^{\epsilon}(0) \to 0$ by the first integral equality. This is a clear contradiction.
Notice that in fact that this really shows that $g' dx = \delta(x)$.
Answer 2: Take $1_{[0,1]}$ instead so that it is an $L^2([0,1])$ function. This is in fact the fourier transform of a "sinc" function, $\sin(k)/k$ up to some normalization constants. If we consider the $H^1$ norm in frequency space we would need $\int_0^{\infty} |k|^2\frac{\sin(k)^2}{|k|^2} < \infty$ which is clearly false. This requires being at ease with the fourier transform so if you're not, answer 1 is probably best.
It is true in $\mathbb{R}^n$ that if $u \in W^{1,p}$ for $p > n$ then $u$ is a.e differentiable and equals a.e its weak gradient (see Evans chapter 5). This is to correct what Tom had said although perhaps we was thinking about the $n=1$ case in which case $2 > 1$.
Hope this helps! Dorian