# Is this equivalent to bounded variation?

Let $\Omega \subset \mathbb{R}^n$ be open and bounded with smooth boundary. For $f \in L^1(\Omega)$, define $$\|D_1 f\|_M(\Omega) =\inf\left\{\liminf_{k\to\infty}\int_\Omega |\nabla f_k|\,dx \mid f_k \to f \text{ in } L^1(\Omega),\ f_k \in \text{ Lip }(\Omega)\right\}.$$ Here $\text{Lip}(\Omega)$ is the set of Lipschitz functions on $\Omega$. Note that by Rademacher's Theorem, for $f \in \text{Lip}(\Omega)$, $\nabla f$ exists Lebesgue-a.e. My question is, is $\|D_1 f\|_M(\Omega)$ the same as $\int_\Omega |Df|$ in general? I have a feeling the answer is ''no'', because if it is ''yes'', people would probably use this as the definition of bounded variation instead of the usual definition, which I find more complicated.

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Is $\omega$ supposed to be a test function? – Jose27 May 28 '12 at 2:51
Oops, the $\omega$ doesn't belong there. I removed it. Thanks. – Stefan Smith May 28 '12 at 7:12