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Let $\phi:\mathbb{R}^2\to\mathbb{R}$ such that $\phi^{-1}(0)\neq\emptyset$ and $\phi\in C^1(W)$ where $W$ is a compact neighborhood of $\phi^{-1}(0)$, with $\nabla\phi\neq 0$ in $W$. So there is some $t_0>0$ such that for all $|t|<t_0$ we have that the set $\{\phi=t\}\subset W$ (even compact inclusion can be assumed). How can we prove that for any $f\in L^1(\mathbb{R}^2)$ the following relation holds:

$$\lim\limits_{t\to 0} \int_{\{\phi=t\}} f(x)\ d\mathcal{H}^1=\int_{\{\phi=0\}} f(x)\ d\mathcal{H}^1 $$

?

I found this result proved in the appendix of the article of G.H.Cottet, E. Maitre - A level set method for fluid-structure interaction with immersed surface (that can be downloaded from here for free https://www.researchgate.net/publication/229059739_A_level_set_method_for_fluid-structure_interactions_with_immersed_surfaces). But the proof uses a $C^2$ regularity for $\phi$. The idea of the proof used there (to generate a family of diffeomorphisms that push $\phi=t$ to $\phi=0$ using orthogonal trajectories) cannot be extended to $C^1$ regularity. The same idea is posted also on MS:https://math.stackexchange.com/questions/505332/parametrization-of-level-sets-of-a-smooth-function

However this result seems to be a consequence of the inverse function theorem. I made a similar post here: Continuity of Hausdorff measure on level sets. The solution posted there has a problem: the inverse function theorem cannot be applied in all points of $\phi=0$ (only on those $x\in\phi^{-1}(0)$ in which $\dfrac{\nabla\phi(x)}{|\nabla\phi (x)|}\neq\pm (0,1)$).

Maybe the following question can be useful Continuity of $r\mapsto\int_{\Sigma\cap B_r(x)}f^2d\mu$.

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    $\begingroup$ I'd say this is essentially Fubini in charts. $\endgroup$
    – Pietro Majer
    Commented Aug 14, 2021 at 8:04
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    $\begingroup$ Sure, but don't you need e.g. $f$ continuous? Otherwise the trace of $f$ on the level set is either not defined (if you think $f$ as defined a.e. or as a class) or possibly discontinuous (if $f$ is a point-wise defined measurable function. $\endgroup$
    – Pietro Majer
    Commented Aug 14, 2021 at 15:00
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    $\begingroup$ For instance consider the simple case of a projection $\phi:(x,y)\mapsto y$, so the level set is the plane $\{y=t\}$, and $f\in L^1([0,1]\times[-1,1])$ is $f(x,y):=\chi_{\mathbb{R}_+}(y)$. $\endgroup$
    – Pietro Majer
    Commented Aug 14, 2021 at 15:03
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    $\begingroup$ For the proof first do the local case where the level sets are graphs: $\{\phi=t\}=\text{graph}(\psi(\cdot,t))$ for some $C^1$ diffeo $$\psi:\Omega\times ]-\epsilon,\epsilon[\to U:=\psi(\Omega\times ]-\epsilon,\epsilon[) $$ for an open set $\Omega\subset\mathbb{R}^{n-1}$. Then for $f\in C_c(U)$, $\int_{\{\phi=t\}}fd\mathcal H^{n-1}=\int_{\mathbb{R}^{n-1}}f(\psi(x,t))\sqrt{|\nabla_x\psi(x,t)|^2+1}dx$ is continuous wrto $t$ by dominated convergence. The general case follows by the implicit function theorem and standard argument by partition of unity. $\endgroup$
    – Pietro Majer
    Commented Aug 14, 2021 at 15:57
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    $\begingroup$ Alternatively you can modify the proof in the link: instead of the normal field, which in your case is only $C^0$, use any Lipschitz or even smooth field transverse to the level sets. You can build it patching together constant local fields by means of a partition of unity. $\endgroup$
    – Pietro Majer
    Commented Aug 14, 2021 at 16:05

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