Proof of Norm Bound for a Vector-Valued Lebesgue Integral over a Compact Interval

Measurability and boundedness of $|g|$. The Euclidean norm satisfies $\big||x|-|y|\big|\le|x-y|$ for $x,y\in\mathbb{R}^n$, by the triangle inequality, so $x\mapsto|x|$ is sequentially continuous on $\mathbb{R}^n$. Since the components of $g$ are measurable, sequentially continuous functions of measurable Euclidean maps shows that $s\mapsto|g(s)|$ is measurable. If $|g^i(s)|\le K_i$ for all $s$ and $i$, then $|g(s)|^2=\sum_i g^i(s)^2\le\sum_iK_i^2$, so $|g|$ is bounded; being nonnegative, bounded and measurable on a set of finite measure, it is integrable and $\int_{[a,b]}|g(s)|\,ds\ge0$ by monotonicity of the integral.

The inequality. If $I=0$ then $|I|=0$ and the inequality holds. Assume $I\neq0$. By linearity of the integral applied to the finite linear combination $\sum_iI^ig^i$,

$$|I|^2=I\cdot I=\sum_{i=1}^nI^i\int_{[a,b]}g^i(s)\,ds=\int_{[a,b]}\Big(\sum_{i=1}^nI^ig^i(s)\Big)ds=\int_{[a,b]}I\cdot g(s)\,ds ,$$

where $|I|^2=I\cdot I$ by the elementary properties of the Euclidean norm. For every $s$, the Cauchy-Schwarz inequality, applied to the positive semidefinite quadratic form given by the identity matrix, whose associated bilinear form is the dot product, gives $I\cdot g(s)\le|I|\,|g(s)|$. By monotonicity and linearity of the integral,

$$|I|^2\le\int_{[a,b]}|I|\,|g(s)|\,ds=|I|\int_{[a,b]}|g(s)|\,ds .$$

Dividing by $|I|>0$ gives $|I|\le\int_{[a,b]}|g(s)|\,ds$. $\blacksquare$