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I am not sure if this is already posted, though, I hope I can get some help, and thank in advance. This question arises from the proof of the following.

Proposition: Let G be a topological group, of which H is a subgroup. Then, H is closed in G, if, and only if, there exists a neighborhood U of 1 in G, such that the intersection of U and H is closed in G.

During the proof, there is an assumption that confuses me, that is, it assumes a neighborhood V of 1, such that V=$V^{\iota}$, and that V*V is a subset of U. But, as far as I am concerned, there is no use of the assumption that $V^{\iota}$=V, and this is exactly my question.

Is it true that we always can find, for every neighborhood U of 1 in G, a neighborhood V of 1 in G such that V=$V^{\iota}$, and that V*V is a subset of U?

I know that the latter assumption results from the continuity of the multiplication, but I am very confused by the former. Thanks and regards here.

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    Dear Theo Buehler: Sorry for the absence recently, I was occupied by something. Although I not yet know what the Radon measures are, I thought that might be what I am trying to understand; moreover, the link in your post is indeed interesting, and I am ready to refresh my mind! Thank you very much. Also, best regards here.2011-07-20

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For the sake of having an answer: Choose a neighborhood $W$ of $1$ such that $W\cdot W \subset U$. This is possible by continuity of the multiplication, as you say. Then $V = W \cap W^{-1}$ is symmetric, i.e., $V^{-1} = V$ and $V\cdot V \subset W \cdot W \subset U$. As inversion is continuous and preserves the identity, $V$ is a neighborhood of $1$.