We prove that Krivine’s Function Calculus is compatible with integration. Let
$(\unicode[STIX]{x1D6FA},\unicode[STIX]{x1D6F4},\unicode[STIX]{x1D707})$ be a finite measure space,
$X$ a Banach lattice,
$\mathbf{x}\in X^{n}$, and
$f:\mathbb{R}^{n}\times \unicode[STIX]{x1D6FA}\rightarrow \mathbb{R}$ a function such that
$f(\cdot ,\unicode[STIX]{x1D714})$ is continuous and positively homogeneous for every
$\unicode[STIX]{x1D714}\in \unicode[STIX]{x1D6FA}$, and
$f(\mathbf{s},\cdot )$ is integrable for every
$\mathbf{s}\in \mathbb{R}^{n}$. Put
$F(\mathbf{s})=\int f(\mathbf{s},\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D707}(\unicode[STIX]{x1D714})$ and define
$F(\mathbf{x})$ and
$f(\mathbf{x},\unicode[STIX]{x1D714})$ via Krivine’s Function Calculus. We prove that under certain natural assumptions
$F(\mathbf{x})=\int f(\mathbf{x},\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D707}(\unicode[STIX]{x1D714})$, where the right hand side is a Bochner integral.