Bourbaki

Example 2

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

Integration, chapter III, § 4. Products of measures

Let us take X=Y=RX = Y = \mathbf{R}, and for λ\lambda and μ\mu the Lebesgue measure on R\mathbf{R} (§1, No. 3); their product is called Lebesgue measure on R2\mathbf{R}^2; the integral of a function f∈K(R2;C)f \in \mathscr{K}(\mathbf{R}^2; \mathbf{C}) with respect to this measure is denoted ∬f(x,y) dx dy\iint f(x, y)\, dx\, dy or ∬f(x,y) dy dx\iint f(x, y)\, dy\, dx; formula (4), for a function that is zero outside a compact rectangle [a,b]×[c,d][a, b] \times [c, d], yields the formula

∫cddy∫abf(x,y) dx=∫abdx∫cdf(x,y) dy \int_c^d dy \int_a^b f(x, y)\, dx = \int_a^b dx \int_c^d f(x, y)\, dy

proved in FRV, II, §3, No. 6.

Since Lebesgue measure on R\mathbf{R} is invariant under every translation (§1, No. 3), it follows at once that Lebesgue measure on R2\mathbf{R}^2 is invariant under every translation of R2\mathbf{R}^2.

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

Cited by

Nothing.

Cites

  • FRV, II, §3, No. 6 not yet on this site: FVR II
  • §1, No. 3
Transcribed from the printed volume. Not translated.