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Let $G$ be a locally compact group with left Haar measure $\mu$.

Let $\delta:G\to(0,\infty)$ the unique modular function such that $\delta(x)\mu(E) = \mu(Ex)$ for all Borel sets $E$ and $x\in G$.

How is this definition used in integration?

Are the following steps correct? (Please alert me of even the slightest mistake!)

For fixed $x\in G$, and Borel set $E$, we have $\delta(x)\mu(E) = \mu(Ex)$.

$\int_{G}f(y)\mu(dy) = \int_{G}f(yx)\mu(dy\cdot x) = \int_{G}\delta(x)f(yx)\mu(dy)$

For a $\mu$-integrable function $f$, I write $\int_{G}f(g)\mu(dy)$ to be mean the same as $\int_{G}fd\mu$, with the indication that $y$ is my variable of integration.

By $\mu(dy\cdot x)$ in the middle step, I mean I am integrating with respect to the measure $\lambda$ defined by $\lambda(E) = \mu(Ex)$.

My motivation is that I am trying to solve this problem: Verifying Convolution Identities

$\bf{\text{Next}}$:

When trying to verify the Convolution identity

$$(f*g)(x) := \int_{G}f(y)g(y^{-1}x)\mu(dy) = \int_{G}\delta(y)^{-1}f(y^{-1})g(yx)\mu(dy)$$

How do I deal with the fact that the argument of $\delta$ is the variable of integration??

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roo
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1 Answers1

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Yes, this looks correct to me. It's how you handle right translation when you've got a left Haar measure. Usually the next step would be to pull the $\delta(x)$ out of the integral.

Ted
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  • Thanks so much! I added an edit with a followup question, which is the main motivation for the question.

    I will accept your answer I just didn't want the question to look like it was totally done and be ignored. :>

     – roo May 12 '13 at 20:01