= Bounded-variation contraction under clipping
Let $T(t)=\min(b,\max(a,t))$, $a\leq b$, and let $u\in BV(\Omega)$. The variation contraction
$$
|D(T\circ u)|(\Omega)\leq|Du|(\Omega)
$$
follows from the one-Lipschitz scalar composition rule, by smooth approximation and <lower semicontinuity>. To conclude global membership $T\circ u\in BV(\Omega)$, one must also retain its <L1 norm>.
That integrability is automatic when $\Omega$ has finite <Lebesgue measure>. It also holds on an arbitrary domain when $T(0)=0$, equivalently $0\in[a,b]$, since $|T(u)|\leq|u|$. Without either condition it can fail: $u=0$ on $\mathbb R$ belongs to <BV space>, but clipping it to $[1,2]$ gives the nonintegrable constant one. Its derivative variation is zero, which does not repair the missing global integrability.
For <total variation denoising> on a bounded domain with data in $[a,b]$, clipping cannot increase either the variation penalty or the pointwise squared error. Uniqueness of the minimizer therefore forces range preservation. The variation inequality and this integrability qualification describe different parts of the argument.
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