Limit of an integral of gradient Let $U$ be an open connected set with smooth boundary in $\mathbb{R}^{d}$ and $U_{n} = \{x \in U: d(x, \partial U) > 1/n\}$. Let $f_{n} \in C_{c}^{\infty}(\mathbb{R}^{d})$ such that $f_{n}(x) = 1$ on $U_{n}$ and $0$ outside $U$ (we can find such an $f_{n}$ by Urysohn's Lemma). Is there a way to rigorously justify that $$\int_{\mathbb{R}^{d}}|\nabla f_{n}|\, dx \rightarrow m(\partial U)$$ as $n \rightarrow \infty$? Here $m$ denotes $d - 1$ dimensional Lebesgue measure.
 A: This is actually not true if you choose $f_n$ in an arbitrary way. For example, let $d=1$ and $U = (-1,1)$. Then consider $f=f_n$ to be smooth and of compact support in $(-1,1)$ such that 


*

*$f(x) = f(-x)$, 

*$f(x) = 1 \text{ for }|x|\leq 1-1/n$, 

*$f(1-1/2n) = 2$,

*$f$ is increasing in $(1-1/n, 1-1/2n)$ and

*$f$ is decreasing in $(1-1/2n, 1)$.


Then 
$$\int_{-1}^1 |f'| dx = 2\int_0^1 |f'| dx = 2\bigg( \int_{1-1/n}^{1-1/2n} f'(x)dx - \int_{1-1/2n}^{1} f'(x)dx\bigg)$$
$$=2\big( f(1-1/2n) -f(1-1/n) - f(1) + f(1-1/2n)\big) = 2 (2-1 -0 + 2) = 6$$
$$\neq 2 = \mathscr H^0 (\partial U)\ .$$
After some thought, one can only at best show:
Theorem We have 
$$\liminf_{n\to \infty} \int_U |\nabla f_n | dx \geq \mathscr H^{d-1}(\partial U)\ .$$
To prove this theorem, first we break $\partial U$ into small portions (called $M$) so that we assume there is a $d-1$ open submanifold $M$ in $\mathbb R^d$ and there is a open set $W$ in $\mathbb R^d$ which admits a parametrization of the normal bundle of $M$. 
That is, $M$ is given by a smooth embedding $F:D\to \mathbb R^d$, $ D \subset \mathbb R^{d-1}$, then locally there is a parametrization of $W$ given by $\tilde F : D \times (-\epsilon, \epsilon)\to W \subset \mathbb R^d$,
$$\tilde F(y^1, \cdots, y^{d-1}, t) = F(y^1, \cdots, y^{d-1}) +t v (y^1, \cdots, y^{d-1}),$$
where $v$ is the normal vector of $M$. Let $f_n$ be a smooth function on $W$, such that 
$$f(y, t) = 0 \text{ when }t\leq 0 \text{ and } f(y, t) = 1\ \text{when }t\geq 1/n\ .$$
Let $g_{ij}$ be the metric on $M$, then on $W$ we have the metric $\tilde g_{ij}$ given by 
$$\tilde g_{ij} = g_{ij}+ t^2 \langle v_i, v_j\rangle \ \text{ for }i, j=1, \cdots ,d-1\ , \tilde g_{tj} = 0, \ \tilde g_{tt} = 1\ .$$
With respect to this parametrization, locally we have 
$$\int_W |\nabla f_n| dx = \int_{-\epsilon}^\epsilon \int_D |\nabla f_n| dV, $$
where 
$$dV = \sqrt{\det(\tilde g_{ij})} dy^1\cdots dy^{n-1}dt \ .$$
On the other hand, decomposes $\nabla f$ into normal and tangential component (corresponds to where $t$ is fixed) $\nabla f = \partial _t + \nabla^\perp f$, then
$$|\nabla f|^2 = |\partial_t f|^2  + |\nabla^\perp f|^2\Rightarrow |\nabla f|\geq |\partial_t  f|\ . $$
$$\Rightarrow \int_W |\nabla f| dx \geq \int_W |\partial_t f| dx \geq  \int_W \partial_t f dx $$
So the theorem follows from the
Claim We have
$$\lim_{n\to \infty} \int_W \partial_t f dx \geq \mathscr H^{d-1}(\partial U)\ .$$
First we have the equality
$$\partial _t f\sqrt{ \det(\tilde g_{ij})} = \partial_t \big(f  \sqrt{ \det(\tilde g_{ij})} \big) - \frac{t}{2} f \sqrt{\det (\tilde g_{ij})} |A|_t^2, $$
where $|A|^2_t$ is the norm of the second fundamental form with respect to the metric $\tilde g_{ij}$ at time $t$. Then using fundamental theorem of calculus, 
$$\int_W \partial_t f dx = \int_D \int_{-\epsilon}^\epsilon \partial _t \sqrt{\det (\tilde g_{ij})} dydt = \int_D dV_t - \frac{t}{2} \int_W f |A|^2_t dx$$
$$= \mathscr H^{d-1} (M_t) - \frac{1}{2} \int_W f t|A|^2_t dx \to \mathscr H^{d-1}(M)$$
as $n\to \infty$. 
(Well, this is a bit messy, please feel free to ask if you do not understand the notations)
Remark So from the proof we see that in order that the statement claimed by the OP is true, we need a good approximation sequence $f_n$ such that they are 


*

*"decreasing along the normal direction" (So that $\partial _t f = |\partial _t f|$), and

*Has "small tangential gradient" $\nabla^\perp f$. 


I guess such a sequence of functions can be constructed, for example for modifying the distance function to $\partial U$. But in general it is not clear what conditions on $f_n$ would guaranteed that the limits exists.
