Volume correction of a sequence

I study the article of Baldo: Minimal interface criterion for phase transitions in mixtures of Cahn-Hilliard fluids. The main point of the article is to prove a $\Gamma$-convergence theorem result. In this cases, as can be seen in the definition of $\Gamma$ convergence the first part is easy, but the last part is almost always constructive, and things get messy.

In the article, in the proof of the second part of the definition of $\Gamma$-convergence a sequence $(u_\varepsilon) \subset L^1(\Omega \subset \Bbb{R}^N;\Bbb{R}^n)$ ($N$ is not necessarily equal to $n$) which converges to $u$ in $L^1(\Omega;\Bbb{R}^n)$ is constructed, and we know that $\int_\Omega |u_\varepsilon -u| \leq K\varepsilon$. We denote $\eta_\varepsilon = \int_\Omega u_\varepsilon(x)dx -\int_\Omega u(x)dx \in \Bbb{R}^n$. Then we have $|\eta_\varepsilon|\leq K\varepsilon$.

The problem is that the sequence $u_\varepsilon$ must be corrected, hopefully on a small set and preserving continuity, such that $\int_\Omega u_\varepsilon=\int_\Omega u$. To do this, in the article (pages 79-80) the author chooses a ball $B_\varepsilon=B(x_0,\varepsilon^{1/N})$ such that $B_\varepsilon \subset \lbrace u_\varepsilon =\alpha \rbrace \subset\Omega$, and defines

$$v_\varepsilon(x) =\begin{cases} u_\varepsilon(x) & x \in \Omega \setminus B_\varepsilon \newline \alpha+ h_\varepsilon(1- \varepsilon^{-1/N}|x-x_0|) & x \in B_\varepsilon \end{cases}$$ where $h_\varepsilon=-N\omega_{N-1}^{-1}\eta_\varepsilon \varepsilon^{(1-N)/N}$ and $\omega_{N-1}$ is the volume of the $N-1$ dimensional unit ball.

In the article it says that from here it follows that $\int_\Omega v_\varepsilon =\int_\Omega u$, but doing some calculations leads to $$\int_\Omega v_\varepsilon(x)dx=\int_\Omega u_\varepsilon(x)dx+h_\varepsilon\int_{B_\varepsilon}(1-\varepsilon^{-1/N}|x-x_0|)dx= \int_\Omega u_\varepsilon(x)dx +h_\varepsilon C \varepsilon$$ where $C$ is a constant, and we would need that the last term be equal to $\eta_\varepsilon$, but the calculations show that the last integral is of order $\varepsilon$, and cannot cancel the power of $\varepsilon$ in the expression of $h_\varepsilon$. Moreover, the expression of $h_\varepsilon$ looks exactly like the integral was calculated in $N-1$ dimensions, not $N$. Anyway, this could be corrected if the expression of $h_\varepsilon$ wouldn't be used in the sequel of the article in an essential way (page 80 bottom).

It is used in proving that

$$\limsup_{\varepsilon \to 0} \int_{B_\varepsilon}\left[ \varepsilon |\nabla v_\varepsilon|^2+\frac{1}{\varepsilon}W(v_\varepsilon)dx \right]=$$ $$\limsup_{\varepsilon \to 0} \left[ \varepsilon |h_\varepsilon|^2\varepsilon^{-2/N}|B_\varepsilon|+\frac{1}{\varepsilon}\int_{B_\varepsilon}W(\alpha+ h_\varepsilon(1- \varepsilon^{-1/N}|x-x_0|) )dx \right]= 0$$ where $W$ is continuous on $\Bbb{R}^n$ with $W(\alpha)=0$.

The gradient term easily converges to zero for the correct expression of $h_\varepsilon$ (and a pretty wide range of powers of $\varepsilon$ in $h_\varepsilon$), but for the term with $W$ to converge to zero we need $h_\varepsilon \to 0$ or the volume of $B_\varepsilon$ to have order greater than $\varepsilon$ and if my calculations are correct, this cannot happen at the same time.

Since the article is widely cited, and the same formula for $h_\varepsilon$ is used in the article of Luciano Modica: Gradient theory of phase transitions with boundary contact energy, which is again very cited, I guess that the calculations can be corrected, or I am not getting this right.

Is there any mistake in my calculations? If I am correct, then is there another way to correct the integrals of $u_\varepsilon$ such that the limsup remains zero?

I know the question seems long, but I included some of the details for it to be a bit self contained. Anyway, the main issue is the choice of the correction of the integral term.

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