[proofplan] We write the calibration assumption as an equality of conditional expectations with respect to the $\sigma$-algebra generated by the predicted probability $Z$. For a bin $I$, the event $A=\{Z\in I\}$ belongs to $\sigma(Z)$, so the defining property of conditional expectation gives equality of the integrals of $Y$ and $Z$ over $A$. Dividing by the positive probability of $A$ gives equality between the empirical event rate and the average predicted probability in that bin. [/proofplan] [step:Represent the bin as an event measurable with respect to the predicted probability] Let $Z:=\hat p(X)$ denote the predicted-probability random variable. Let $\mathcal B([0,1])$ denote the Borel $\sigma$-algebra on $[0,1]$. Fix a Borel measurable interval $I\subset[0,1]$ with $\mathbb P(Z\in I)>0$. Define the event \begin{align*} A := \{\omega\in\Omega : Z(\omega)\in I\}. \end{align*} Since $Z:(\Omega,\mathcal F)\to([0,1],\mathcal B([0,1]))$ is measurable and $I\in\mathcal B([0,1])$, we have $A=Z^{-1}(I)\in\sigma(Z)$. Also, by hypothesis, $\mathbb P(A)>0$. [/step] [step:Use the defining property of conditional expectation on the bin event] The random variables $Y$ and $Z$ are integrable because $0\leq Y\leq 1$ and $0\leq Z\leq 1$ $\mathbb P$-a.s. Since $A\in\sigma(Z)$, the defining property of $\mathbb E[Y\mid\sigma(Z)]$ gives \begin{align*} \int_A Y\,d\mathbb P(\omega) = \int_A \mathbb E[Y\mid\sigma(Z)]\,d\mathbb P(\omega). \end{align*} Using the calibration assumption $\mathbb E[Y\mid\sigma(Z)]=Z$ $\mathbb P$-a.s., we obtain \begin{align*} \int_A Y\,d\mathbb P(\omega) = \int_A Z\,d\mathbb P(\omega). \end{align*} [guided] The purpose of introducing $A$ is that it describes exactly the observations whose predicted probabilities fall inside the interval $I$. Because $A=Z^{-1}(I)$ and $I$ is Borel measurable, the event $A$ is measurable with respect to $\sigma(Z)$. This is the key measurability point: conditional expectation with respect to $\sigma(Z)$ controls integrals over every event whose occurrence is determined by $Z$. The random variables being integrated are integrable. Indeed, $Y$ takes values in $\{0,1\}$, so $0\leq Y\leq 1$, and $Z$ takes values in $[0,1]$, so $0\leq Z\leq 1$. Therefore \begin{align*} \int_\Omega |Y|\,d\mathbb P(\omega) \leq 1, \qquad \int_\Omega |Z|\,d\mathbb P(\omega) \leq 1. \end{align*} Since $A\in\sigma(Z)$, the defining property of the conditional expectation $\mathbb E[Y\mid\sigma(Z)]$ yields \begin{align*} \int_A Y\,d\mathbb P(\omega) = \int_A \mathbb E[Y\mid\sigma(Z)]\,d\mathbb P(\omega). \end{align*} The calibration hypothesis states that $\mathbb E[Y\mid\sigma(Z)]$ equals $Z$ almost surely. Replacing the conditional expectation by $Z$ inside the integral is valid because changing an integrable random variable on a $\mathbb P$-null set does not change its integral. Hence \begin{align*} \int_A Y\,d\mathbb P(\omega) = \int_A Z\,d\mathbb P(\omega). \end{align*} [/guided] [/step] [step:Divide by the bin probability to obtain the conditional expectation identity] Because $\mathbb P(A)>0$, the conditional expectations given the event $A$ are \begin{align*} \mathbb E[Y\mid A] &= \frac{1}{\mathbb P(A)}\int_A Y\,d\mathbb P(\omega),\\ \mathbb E[Z\mid A] &= \frac{1}{\mathbb P(A)}\int_A Z\,d\mathbb P(\omega). \end{align*} The integral equality from the previous step therefore implies \begin{align*} \mathbb E[Y\mid A] = \mathbb E[Z\mid A]. \end{align*} Since $A=\{Z\in I\}$, this is exactly \begin{align*} \mathbb E[Y\mid Z\in I] = \mathbb E[Z\mid Z\in I]. \end{align*} Finally, substituting $Z=\hat p(X)$ gives \begin{align*} \mathbb E[Y\mid \hat p(X)\in I] = \mathbb E[\hat p(X)\mid \hat p(X)\in I], \end{align*} which proves the claim. [/step]