Proof of Square-Integrable Vector-Valued Functions: Components, Dot Products and Almost-Everywhere Equality

Pointwise identities. Let $f,g:X\to\mathbb{R}^{n}$ be functions whose components are all measurable. For $x\in X$, by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (i) and The Dot Product on $\mathbb{R}^n$ (i),

$$|f|^{2}(x)=\lVert f(x)\rVert^{2}=\sum_{k=1}^{n}f_{k}(x)^{2}=\sum_{k=1}^{n}|f_{k}|^{2}(x),\qquad (f\cdot g)(x)=\sum_{k=1}^{n}f_{k}(x)\,g_{k}(x),$$

where $|f_{k}|^{2}$ is the function of Power-Integrable Functions and the p-Seminorm (i) with $p=2$, whose value $|f_{k}(x)|^{2}$, a real power, is $f_{k}(x)^{2}$ by Properties of Real Powers (i). Since the terms are nonnegative, Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (vi) gives

$$|f_{k}|^{2}(x)\le|f|^{2}(x)\qquad(x\in X,\ k\in[n]). \tag{1}$$

Components. Let $f:X\to\mathbb{R}^{n}$. If $f\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$, then $f$ is measurable and so is every $f_{k}$; if every $f_{k}$ lies in $\mathcal{L}^{2}(\mu)$, every $f_{k}$ is measurable and so is $f$ (both by Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (i)). In either case the pointwise identities apply.

Suppose $f\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ and let $k\in[n]$. By (1) and Basic Properties of the Integral of Nonnegative Measurable Maps (iii), $\int|f_{k}|^{2}\,d\mu\le\int|f|^{2}\,d\mu<\infty$, so $f_{k}\in\mathcal{L}^{2}(\mu)$ (Power-Integrable Functions and the p-Seminorm (ii)).

Suppose now $f_{k}\in\mathcal{L}^{2}(\mu)$ for every $k\in[n]$. By Elementary Properties of the p-Seminorm (ii), $|f_{k}|^{2}\in\mathcal{L}^{1}(\mu)$ with $\int|f_{k}|^{2}\,d\mu=\|f_{k}\|_{2}^{2}$. By Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (viii), with all coefficients $1$, the function $\sum_{k=1}^{n}|f_{k}|^{2}$, which is $|f|^{2}$, lies in $\mathcal{L}^{1}(\mu)$ and

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

As $|f|^{2}$ is measurable with nonnegative real values, Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi) shows that its integral as an element of $\mathcal{M}^{+}$ is finite and equal to this one. Hence $f\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$, and the formula of (i) holds. By the previous paragraph it holds for every $f\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$.

Dot products. Let $f,g\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$. By (i), $f_{k},g_{k}\in\mathcal{L}^{2}(\mu)$, so $f_{k}g_{k}\in\mathcal{L}^{1}(\mu)$ for every $k\in[n]$ by Hölder's Inequality (ii). As $f\cdot g=\sum_{k=1}^{n}f_{k}g_{k}$ pointwise, Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (viii) gives $f\cdot g\in\mathcal{L}^{1}(\mu)$ and the formula of (ii).

Linear combinations. Let $f,g\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ and $a,b\in\mathbb{R}$. For $k\in[n]$, $(af+bg)_{k}=af_{k}+bg_{k}$ by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv), and this lies in $\mathcal{L}^{2}(\mu)$ by (i) and Minkowski's Inequality (i) with $p=2$. By (i), $af+bg\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$. By the vector-space laws of $\mathbb{R}^{n}$ at each point, $f+g=1f+1g$, $af=af+0g$, $-f=(-1)f+0g$ and $f-g=1f+(-1)g$, which gives the particular cases.

Equivalence relation. For functions $f,g:X\to\mathbb{R}^{n}$ write $E(f,g)=\{x\in X:f(x)=g(x)\}$, so that $f\sim g$ means that $E(f,g)$ holds $\mu$-almost everywhere. Let $f,g,h\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$. Reflexivity: $X\setminus E(f,f)=\emptyset$, which is $\mu$-null since $\emptyset\in\mathcal{F}$ and $\mu(\emptyset)=0$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vi), Measure, Measure Space, and Probability Measure (i) and Null Sets and Properties Holding Almost Everywhere (i)). Symmetry: $E(f,g)=E(g,f)$. Transitivity: if $f\sim g$ and $g\sim h$, then $E(f,g)\cap E(g,h)$ holds almost everywhere by Null Sets and Properties Holding Almost Everywhere (vii), and it is contained in $E(f,h)$, so $f\sim h$ by Null Sets and Properties Holding Almost Everywhere (vi). Thus $\sim$ is an equivalence relation on $\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ (Equivalence Relations on a Set (i)).

If $f\sim g$ and $k\in[n]$, then $E(f,g)\subseteq\{x\in X:f_{k}(x)=g_{k}(x)\}$, so $f_{k}=g_{k}$ almost everywhere by Null Sets and Properties Holding Almost Everywhere (vi). Conversely, suppose $f_{k}=g_{k}$ almost everywhere for every $k\in[n]$. By (iii), $f-g\in\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$, and its components $(f-g)_{k}=f_{k}-g_{k}$ (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv)) lie in $\mathcal{L}^{2}(\mu)$ by (i) and vanish almost everywhere, so $\|f_{k}-g_{k}\|_{2}=0$ by Elementary Properties of the p-Seminorm (vii). By (i) and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (iv),

$$\int|f-g|^{2}\,d\mu=\sum_{k=1}^{n}\|f_{k}-g_{k}\|_{2}^{2}=0 .$$

By The Lebesgue Integral and Null Sets (iv), $\lVert f(x)-g(x)\rVert^{2}=0$ for almost every $x$, and at every such $x$, $f(x)=g(x)$ by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii). Hence $f\sim g$, by Null Sets and Properties Holding Almost Everywhere (vi).

Compatibility. Let $f,f',g,g'$ and $a$ be as in (v). The set $Q=E(f,f')\cap E(g,g')$ holds almost everywhere by Null Sets and Properties Holding Almost Everywhere (vii), and for $x\in Q$,

$$f(x)+g(x)=f'(x)+g'(x),\qquad af(x)=af'(x),\qquad f(x)\cdot g(x)=f'(x)\cdot g'(x).$$

By Null Sets and Properties Holding Almost Everywhere (vi), $f+g\sim f'+g'$ and $af\sim af'$, these functions lying in $\mathcal{L}^{2}(\mu;\mathbb{R}^{n})$ by (iii), and $f\cdot g=f'\cdot g'$ almost everywhere. Since $f\cdot g$ and $f'\cdot g'$ lie in $\mathcal{L}^{1}(\mu)$ by (ii), their integrals agree by The Lebesgue Integral and Null Sets (iii).