Let $\gamma_{1}=\mathscr{N}(0,I_{1})$ in $\mathbb{R}$ be the standard Gaussian measure. Consider the sequence $(f_{n})_{n\in\mathbb{N}}\in C_{b}^{1}(\mathbb{R})$ defined by $$f_{n}(x)=\begin{cases} 0,&\mbox{for }x\le 0, \\ nx,&\mbox{for }0\le x \le\frac{1}{n}, \\ 1,&\mbox{for }x\ge\frac{1}{n}. \end{cases}$$
I have already proven that $f_{n}\to f$ in $L^{1}(\mathbb{R},\gamma_{1})$ and that $\sup_{n\in\mathbb{N}}\|f_{n}\|_{W^{1,1}(\mathbb{R},\gamma_{1})}<\infty$, but now I want to prove that $f\notin W^{1,1}(\mathbb{R},\gamma_{1})$.
For every $f\in W^{1,1}(\mathbb{R},\gamma_{1})$, the weak derivative $Df$ must coincide with $\partial f/\partial x$. So I want to prove that this is not the case for our $f$.
We have $f(x)=0$ on the negative half-plane and $f(x)=1$ on the positive half-plane. So we can write $f(x)=1_{\Gamma}(x)$, with $\Gamma$ a positive half line. Then $\partial f/\partial x=0$ since the Dirac function is a generalised function.
$f$ has a weak derivative $Df$ if $$\int_{\mathbb{R}}f\frac{\partial g}{\partial x}d\gamma_{1}=-\int_{\mathbb{R}}g\frac{\partial f}{\partial x}d\gamma_{1}+\int_{\mathbb{R}}f(x)g(x)x\gamma_{1}(dx)$$ holds $\forall g\in C_{b}^{1}(\mathbb{R})$.
Do I now just substitute my $\partial f/\partial x=0$?
In which case, we would have $$\int_{\mathbb{R}}f\frac{\partial g}{\partial x}d\gamma_{1}=\int_{\mathbb{R}}f(x)g(x)x\gamma_{1}(dx)$$
There is a Lemma which states that
For every $f\in C^{1}_{b}(\mathbb{R})$ $$\int_{\mathbb{R}}\frac{\partial f}{\partial x}(x)\gamma_{1}(dx)=\int_{\mathbb{R}}x f(x)\gamma_{1}(dx)$$
Can I apply this somehow to get what I want?
Remember that $f \in W^{1,1} (\gamma_1)$ means that $f, f' \in L^1 (\gamma_1)$ where $f'$ is understood in the weak sense. Remember also that $f'$ in the Sobolev sense in the same as $f'$ in the distributional sense, therefore let us compute the latter one. Finally, we shall use that $\gamma_1 = \dfrac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2}$.
Note, first, that $f = \left\{ \begin{eqnarray} 0 &,& x \le 0 \\ 1 &,& x > 0 \end{eqnarray} \right.$.
Pick $\varphi$ smooth with compact support, arbitrary. We want to compute $f'$ (as a distribution) such that
$$\int f' \varphi {\rm d} \gamma_1 = - \int f \varphi' {\rm d} \gamma_1 + \int f \varphi x {\rm d} \gamma_1 = - \int \limits _0 ^\infty \varphi' {\rm d} \gamma_1 + \int \limits _0 ^\infty \varphi x {\rm d} \gamma_1 = \\ - \int \limits _0 ^\infty \varphi' \frac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2} {\rm d} x + \int \limits _0 ^\infty \varphi x \frac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2} {\rm d} x = \\ - \frac 1 {\sqrt {2 \pi}} \left( \left. \varphi {\rm e}^{-\frac {x^2} 2} \right| _0 ^\infty - \int \limits _0 ^\infty \varphi {\rm e}^{-\frac {x^2} 2} (-x) {\rm d} x \right) + \int \limits _0 ^\infty \varphi x \frac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2} {\rm d} x = \\ \frac {\varphi (0)} {\sqrt {2 \pi}} - \frac 1 {\sqrt {2 \pi}} \int \limits _0 ^\infty \varphi {\rm e}^{-\frac {x^2} 2} x {\rm d} x + \int \limits _0 ^\infty \varphi x \frac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2} {\rm d} x = \frac {\varphi (0)} {\sqrt {2 \pi}} .$$
Noting that the leftmost-hand term can be explicited as $\frac 1 {\sqrt{2 \pi}} \int f' \varphi {\rm e}^{-\frac {x^2} 2} {\rm d} x$, if you equate the beginning of the previous chain of equalities to its end, you get
$$\int \limits _0 ^\infty f' {\rm e}^{-\frac {x^2} 2} \varphi {\rm d} x = \varphi (0) = \delta (\varphi)$$
and, since this is true for arbitrary $\varphi$, this implies that $f' {\rm e}^{-\frac {x^2} 2} = \delta$ (as distributions, with $\delta$ being Dirac's distribution), whence it follows that $f' = {\rm e}^{\frac {x^2} 2} \delta$.
Remeber now that, if $g$ is some smooth function, then $g \delta = g(0) \delta$, the argument being that
$$\langle g \delta, \varphi \rangle = \langle \delta, g \varphi \rangle = g(0) \varphi (0) = g(0) \delta (\varphi) .$$
Using this, the previous expression for $f'$ can be rewritten as $\color{blue} {f' = \delta}$, which is not a regular distribution (i.e. it does not come from a locally-integrable function), therefore it has no chance of being in $L^1 (\gamma_1)$ (because everything living in $L^1$ should also be locally-integrable).