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Let $B$ be a Banach space and let $f$ be a strongly measurable function on a measure space. It is Bochner integrable exactly whenStrong measurability means almost-everywhere norm approximation by simple vector-valued functions. The Bochner integral is the norm limit of their integrals when the approximation also tends to zero in the integral norm; the limit is independent of the approximating sequence.^[ref]

For $f(t)=(t,t^2)$ on $[0,1]$ with values in ${\mathbb{R}}^2$, the integral is $(1/2,1/3)$. In an infinite-dimensional Banach space, norm convergence is part of the construction and cannot be replaced by coordinatewise convergence alone. The boundfollows from the triangle inequality for simple functions and passage to the norm limit. A bounded linear operator commutes with the integral by the same argument. Strong measurability alone does not imply integrability: the scalar example $f(t)=1/t$ on $(0,1)$ fails the norm-integral condition.

Choose the common dominating probability measure $\nu =(\mu +{\mu }^{\prime })/2$ and let $p=d\mu /d\nu$, $q=d{\mu }^{\prime }/d\nu$ be the Radon-Nikodym derivatives. The assumed second moments imply first-moment integrability by the Cauchy-Schwarz inequality. Since $Y$ is a separable Banach space, the measurable $f$ is a strongly measurable function, and both expected values exist as Bochner integrals.

The difference of these Bochner integrals satisfiesThe first inequality is the norm bound for a Bochner integral, and the second is the scalar Cauchy-Schwarz inequality. Since $(\sqrt{p}+\sqrt{q}{)}^2\le 2(p+q)$, the first factor is at mostWith the Hellinger distance convention of the original PDF, the second factor is $\sqrt{2}{d}_{\mathrm{Hell}}(\mu ,{\mu }^{\prime })$. Multiplication proves the Hellinger bound for differences of expectations:The proof uses only the two laws and their common dominating measure; no absolute continuity of measures assumption between $\mu$ and ${\mu }^{\prime }$ is needed.