Rademacher's theorem in reflexive spaces. In a course about semigroups of operators my professor used a generalization of Rademacher's theorem:

Let $X$ be a reflexive Banach space. Then every Lipschitz function $f: \ [0,t] \to X$ is differentiable almost everywhere.

I can prove this using the fact that every reflexive Banach space has Radon-Nikodym property. But I am looking for a simpler proof. My prof mentioned that this can be done by applying the one-dimensional Rademacher theorem to functions $ \langle x^*, f(t) \rangle$ but I can't prove it this way. Any help?
 A: Consider a Lipschitz function $f\colon [0,1]\to X$, where $X$ is reflexive. Re-scaling, if necessary, we may assume that $f$ has Lipschitz constant 1. You may also assume that $X$ is separable as $[0,1]$ is separable, hence so it the image of $f$, so we may restrict ourselves to the closed linear span of the image of $f$, which is also reflexive.
Given $y\in X^*$, the function $\omega \mapsto \langle y, f(\omega)\rangle$ is $\|y\|$-Lipschitz, so there is a positive density function $g_y\in L_\infty[0,1]$ with $\|g\|_{L_\infty}\leqslant \|y\|$ and such that
$$\langle y, f(\omega)\rangle = \int\limits_0^\omega g_y(\tau)\,{\rm d}\tau$$
for almost all $\omega$.
As $X$ is separable, so is $X^*$ (because $X$ is reflexive), so we may pick a countable, dense set $D\subset X^*$. For each finite linear combination $y=\sum_{k=1}^n c_k d_k$ where $d_j\in D$ for $j\leqslant n$, we have
$$\langle y, f(\omega)\rangle = \int\limits_0^\omega\left(\sum_{k=1}^nc_k g_{d_k}(\tau)\right)\,{\rm d}\tau$$
for almost all $\omega$, which means that the density of $y$ is the linear combination of densities of $d_k$ with coefficients $c_k$ ($k\leqslant n$). In particular, the essential supremum of $\sum_{k=1}^nc_k g_{d_k}$ is bounded by $\| \sum_{k=1}^n c_k d_k\|$. 
Restricting the coefficients $c_k$ to rational numbers (or to elements of $\mathbb{Q}+i\mathbb{Q}$ in the complex case), we conclude that there are only countably many linear combinations of the form $\sum_{k=1}^nc_k {d_k}$ and they are dense in $X^*$. Consequently, there exists a measure-zero set $\Omega\subset [0,1]$ such that 
$$\left| \sum_{k=1}^nc_k g_{d_k}(\omega) \right| \leqslant \left\| \sum_{k=1}^n c_k d_k\right\|$$
for all $\omega\in [0,1]\setminus \Omega$ and all such linear combinations. By  density, this holds for all linear combinations of elementes from $D$. Therefore, given $\omega\in [0,1]\setminus \Omega$, the map $$y\mapsto g_y(\omega)\quad (y\in X^*)$$ is a well-defined, bounded, linear functional $\psi_\omega$ on $X^*$, so $\psi_\omega\in X^{**}=X$ with $\|\psi_\omega\|\leqslant 1$.
Take $y\in D$. The map
$$\langle y, f(\omega)\rangle = \int\limits_0^\omega \langle y, \psi_\tau\rangle\,{\rm d}\tau$$
is Lebesgue-measurable (and bounded). We are now in a position to apply Lebesgue's dominated convergence theorem to see that the above formula holds for all $y\in X^*$. As $X$ is separable, by the Pettis measurability theorem, the function $\psi(\omega)=\psi_\omega$ is Bochner-measurable. It is also bounded and therefore integrable. Finally, 
$$f(\omega)=\int\limits_0^\omega \psi(\tau)\,{\rm d}\tau$$
for almost all $\omega$. It follows from the fundamental theorem of calculus for the Bochner integral that $f$ is differentiable almost everywhere.
