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Hello,

Prop 11.3. Let G be a topological group.

  1. If G is $T_1$ then G is Hausdorff
  2. If G is not $T_1$, let H be the closure of ${e}$. Then H is a normal subgroup, and if G/H is given the quotient topology, G/H is a Hausdorff topological group.

Proof omitted.

The author then says that it's easy to see that borel measurable functions on G are constant on the cosets of H. Why is this?

References: GB Folland - Real Analysis, Modern techniques and their applications, 2nd ed, chapter 11

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Because every Borel subset of $G$ is a union of cosets of $H$. This is not a research-level question, see the faq. – Emil Jeřábek Mar 2 2012 at 11:45

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