Bourbaki

Exercise 25, § 4

Integration, VIII, p. 71

Let G be a locally compact group, β a left Haar measure on G, μ a positive measure on G. If A is a μ-integrable set and B is a Borel set in G, the function

u:s↦μ(A∩sB) u : s \mapsto \mu(A \cap sB)

is β-measurable in G; if, in addition, B^{-1} is β-integrable, then so is u, and

∫μ(A∩sB) dβ(s)=μ(A)β(B−1). \int \mu(A \cap sB)\ d\beta(s) = \mu(A)\beta(B^{-1}) .

Give an example of a measure μ\mu such that s↦μ(A∩sB)s \mapsto \mu(A \cap sB) is not continuous; if μ\mu is a measure with base β\beta, and if AA is μ\mu-integrable and BB is a Borel set, then the function s↦μ(A∩sB)s \mapsto \mu(A \cap sB) is continuous on GG.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.