TheoremBase

Each clause follows from the fact that a.e. statements are closed under finite intersections and enlargement, together with the a.e. invariance of the seminorm and of the integral.

Proof

For functions u,v:X→Ru,v:X\to\mathbb{R} write

E(u,v)={x∈X:u(x)=v(x)},E(u,v)=\{x\in X:u(x)=v(x)\},

so that, for u,v∈Lp(μ)u,v\in\mathcal{L}^{p}(\mu), u∼vu\sim v means that E(u,v)E(u,v) holds μ\mu-almost everywhere (A Property Holding Almost Everywhere (i)). Two facts about almost-everywhere statements are used throughout: if QQ and RR hold μ\mu-a.e., so does Q∩RQ\cap R, and if QQ holds μ\mu-a.e. and X∩Q⊆RX\cap Q\subseteq R, then RR holds μ\mu-a.e. (Null Sets and Properties Holding Almost Everywhere (vii), Null Sets and Properties Holding Almost Everywhere (vi)).

Equivalence relation. Let f,g,h∈Lp(μ)f,g,h\in\mathcal{L}^{p}(\mu). Since E(f,f)=XE(f,f)=X and X∖X=∅X\setminus X=\emptyset with ∅∈F\emptyset\in\mathcal{F} and μ(∅)=0\mu(\emptyset)=0 (Measure, Measure Space, and Probability Measure (i)), the set X∖E(f,f)X\setminus E(f,f) is μ\mu-null (Null Set of a Measure (i)), so f∼ff\sim f. Since E(f,g)=E(g,f)E(f,g)=E(g,f), f∼gf\sim g implies g∼fg\sim f. Finally

E(f,g)∩E(g,h)⊆E(f,h),E(f,g)\cap E(g,h)\subseteq E(f,h),

so f∼gf\sim g and g∼hg\sim h imply f∼hf\sim h. Thus ∼\sim is reflexive, symmetric and transitive, an equivalence relation (Equivalence Relations on a Set (i)).

Compatibility. The functions f+gf+g, f′+g′f'+g', cfcf and cf′cf' lie in Lp(μ)\mathcal{L}^{p}(\mu) by Minkowski's Inequality (i). Since the operations are pointwise,

E(f,f′)∩E(g,g′)⊆E(f+g,f′+g′),E(f,f′)⊆E(cf,cf′),E(f,f')\cap E(g,g')\subseteq E(f+g,f'+g'),\qquad E(f,f')\subseteq E(cf,cf'),

so f+g∼f′+g′f+g\sim f'+g' and cf∼cf′cf\sim cf'.

The seminorm. The function gg is measurable and g(x)=f(x)g(x)=f(x) for almost every xx, so ∥g∥p=∥f∥p\|g\|_{p}=\|f\|_{p} by Elementary Properties of the p-Seminorm (vi).

The integral of a product. By Hölder's Inequality (ii), fgfg and f′g′f'g' lie in L1(μ)\mathcal{L}^{1}(\mu); in particular f′g′f'g' is measurable. Since

E(f,f′)∩E(g,g′)⊆E(fg,f′g′),E(f,f')\cap E(g,g')\subseteq E(fg,f'g'),

we have fg=f′g′fg=f'g' almost everywhere, and ∫fg dμ=∫f′g′ dμ\int fg\,d\mu=\int f'g'\,d\mu by The Lebesgue Integral and Null Sets (iii).

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