L²(μ;ℝⁿ) Is a Real Hilbert Space

In the settings of Measurable Functions and the Lebesgue Integral: Standing Notation and Real Normed, Inner Product and Hilbert Spaces: Notation, let $(X,\mathcal{F},\mu)$ be a measure space and let $n\in\mathbb{N}$. $L^{2}(\mu;\mathbb{R}^{n})$ and its elements $[f]$ are as in The Lebesgue Space L²(μ;ℝⁿ) (i), with the sum and scalar multiplication of The Lebesgue Space L²(μ;ℝⁿ) (ii) and the inner product $\langle\cdot,\cdot\rangle$ of The Lebesgue Space L²(μ;ℝⁿ) (iii); $|f|^{2}$ and the set $\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ of square-integrable functions are as in Square-Integrable Vector-Valued Functions (i) and Square-Integrable Vector-Valued Functions (ii), and $\|g\|_{2}$, for $g$ in the set $\mathcal{L}^{2}(\mu)$ of Power-Integrable Functions and the p-Seminorm (ii), is the $2$-seminorm. $0$ denotes the function $X\to\mathbb{R}^{n}$ with value the origin, and $-f$ the function $x\mapsto-f(x)$.

(i) Inner product space. $L^{2}(\mu;\mathbb{R}^{n})$, with this sum and scalar multiplication, is a real vector space, with zero vector $[0]$ and negatives $-[f]=[-f]$; and $\langle\cdot,\cdot\rangle$ is an inner product on it, so that $L^{2}(\mu;\mathbb{R}^{n})$ is a real inner product space.

(ii) Norm. For every $f\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$, the induced norm of $[f]$ satisfies

$$\|[f]\|^{2}=\int|f|^{2}\,d\mu=\sum_{k=1}^{n}\|f_{k}\|_{2}^{2}.$$

(iii) Completeness. Assume the axiom of dependent choice. Then $L^{2}(\mu;\mathbb{R}^{n})$ is a real Hilbert space.