Bourbaki

Example

Tag 08LO, permanent. Cite it as [Bourbaki, Tag 08LO].

Integration, chapter III, § 1. Measures on a locally compact space

— 1) The measure εa\varepsilon_a defined by a unit mass at a point a∈Xa \in X is bounded, and ∥εa∥=1\|\varepsilon_a\| = 1.

2) The Lebesgue measure on R\mathbf{R} is not bounded; indeed, for every integer n>0n > 0 there exists a function f∈K(R;C)f \in \mathscr{K}(\mathbf{R}; \mathbf{C}) with values in [0,1][0, 1] and equal to 1 on the interval [−n,n][ -n, n ] (No. 2, Lemma 1); thus ∥f∥=1\|f\| = 1 and

∫−∞+∞f(x) dx≥∫−nnf(x) dx=2n, \int_{-\infty}^{+\infty} f(x)\, dx \geq \int_{-n}^{n} f(x)\, dx = 2n ,

which proves that there does not exist any finite number M satisfying the relation (19).

3) On the real line R\mathbf{R}, the mapping

f↦∫−∞+∞f(x) dx1+x2 f \mapsto \int_{-\infty}^{+\infty} \frac{f(x)\,dx}{1 + x^2}

is a bounded measure since, for every function f∈H(R;C)f \in \mathscr{H}(\mathbf{R}; \mathbf{C}),

∣∫−∞+∞f(x) dx1+x2∣≤∥f∥∫−∞+∞dx1+x2=π⋅∥f∥. \left| \int_{-\infty}^{+\infty} \frac{f(x)\,dx}{1 + x^2} \right| \leq \|f\| \int_{-\infty}^{+\infty} \frac{dx}{1 + x^2} = \pi \cdot \|f\|.

Since the relations ∥f∥≤1\|f\| \leq 1 and ∥f‾∥≤1\|\overline{f}\| \leq 1 are equivalent, it follows from (20) that

(21)

∥μ‾∥=∥μ∥ \|\overline{\mu}\| = \|\mu\|
for every measure μ\mu on XX.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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Cites

Transcribed from the printed volume. Not translated.