By the properties of the Monge-Ampère measure (Proposition 3.93) and of the Legendre-Fenchel dual (Proposition 3.18) the left-hand side is continuous with respect to uniform convergence of functions. Again by Proposition 3.18 and the discussion before the lemma, the right-hand side is also continuous with respect to uniform convergence of functions. Therefore it is enough to treat the case when $\psi$ is smooth and strictly concave. Then

$$2\int_{\mathbb{R}}\psi^{\vee}\circ\partial\psi\,{\mathcal{M}}_{\mathbb{Z}}(\psi)=2\int_{\mathbb{R}}(\psi(u)-u\psi^{\prime}(u))\psi^{\prime\prime}(u)\,\text{\rm d}u.$$

Consider the function

	$\displaystyle\gamma(u)$	$\displaystyle=(\psi^{\prime}(u)-\frac{a+b}{2})\psi(u)-u\frac{(\psi^{\prime})^{2}}{2}+u\frac{ab}{2}$
		$\displaystyle=-(\psi^{\prime}(u)-\frac{a+b}{2})\psi^{\vee}(\psi^{\prime}(u))-\frac{u}{2}(\psi^{\prime}(u)-a)(b-\psi^{\prime}(u)).$

Then

$$\lim_{u\to\infty}\gamma(u)=\frac{b-a}{2}\psi^{\vee}(a),\qquad\lim_{u\to-\infty}\gamma(u)=\frac{a-b}{2}\psi^{\vee}(b),$$

and

$$\,\text{\rm d}\gamma=(\psi-u\psi^{\prime})\psi^{\prime\prime}\,\text{\rm d}u-\frac{1}{2}(\psi^{\prime}-a)(b-\psi^{\prime})\,\text{\rm d}u,$$

from which the result follows. ∎