Measure of $\Omega_k = \{x\in \Omega\, : \, \operatorname{dist}(x,\partial \Omega)>\frac{1}{k}\}$, $\Omega$ is a bounded domain with smooth boundary

Question

I'm trying to prove the following:

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ with smooth boundary. By considering $\Omega_k = \{x\in \Omega\, : \, \operatorname{dist}(x,\partial \Omega)>\frac{1}{k}\}$ , show that $C^{\infty}_c(\Omega)$ (smooth with compact support) is dense in $L^p(\Omega)$.

I think the point is to take $f\in L^p(\Omega)$ and to consider $f_k = f\chi_k$, where $\chi_k$ is the indicator function on $\Omega_k$. Then if $f_k \rightarrow f$ in $L^P(\Omega)$, I can show the result via mollification etc.

My question: Is it true that $f_k \rightarrow f$ in $L^P(\Omega)$ ? In other words: $$\int_{\Omega/\Omega_k} |f|^p dx \rightarrow 0$$ $\Omega/\Omega_k = \{x\in \Omega\, : \, \operatorname{dist}(x,\partial \Omega)\leq\frac{1}{k}\}$ and intuitively it seems that this set tends to a set of measure $0$, and so I can conclude via DCT. But i'm struggling to make this intuitive idea rigorous.

I'm currently reading Adams Sobolev spaces and i'm seeing sets of this form being used often in proofs and intuitively it seems that the point of sets similar to these is to approximate functions inside these sets nicely and outside doesn't matter much because we should be able to make their measures small enough. (eg. theorem 2.21,3.17 in adams but in Adams $\Omega$ is open and not necessarily bounded).

2. What is the general idea behind considering these sets ?

There might be some basic facts about lebesgue measure and or compact sets in $\mathbb{R}^n$ that I have forgotten which imply the convergence above and these would also be very helpful.

Answer 1

Actually, one can use an abstract argument based on the monotone convergence theorem. Indeed, $$ \lambda(\Omega \setminus \Omega_k) = \int_\Omega \chi_{\Omega \setminus \Omega_k} \, \mathrm{d}x \to 0,$$ since the integrand is monotonely decreasing and converges pointwise everywhere to $0$.

Answer 2

In case anyone got confused as I did - The domain is assumed to be an Open subset of $\mathbb{R}^n$. (Sometimes even connected). Moral of the story - Check the definition of a Domain.

So if $\Omega$ is open, then $\chi_{\Omega/\Omega_k}$ converges pointwise to $0$ in $\Omega$, and hence via MCT or DCT $f_k \rightarrow f$ in $L^p(\Omega)$