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Note: I don't yet have a solution to my main issue yet which I have elaborated on in the edit. Further attention is deeply appreciated. :>

$\bf{\text{Original Question}}$:

Let $G$ be a locally compact group, $\mu$ a Haar measure for $G$ and $f,g\in L^{1}(G)$.

Then by definition $(f*g)(x) = \int_{G}f(y)g(y^{-1}x)\mu(dy)$.

The author remarks in the book I am reading that we also have the following identities:

\begin{eqnarray*} (f*g)(x) &=& \int_{G}f(xy)g(y^{-1})\mu(dy)\\ &=& \int_{G}\delta(y^{-1})f(y^{-1})g(yx)\mu(dy)\\ &=& \int_{G}\delta(y^{-1})f(xy)g(y)\mu(dy) \end{eqnarray*}

where $\delta:G\to(0,\infty)$ is the unique modular function such that $\delta(x)\mu(B) = \mu(Bx)$ for all Borel sets $B$ and $x\in G$.

I cannot seem to get verify any of them. Can anyone help me get started on this? The smallest suggestion possible is best.

I was trying to use the property that $\int_{G}f(xy)\mu(dy) = \int_{G}f(y)\mu(x\cdot dy)$ but I couldn't get anywhere. Any suggestions on what the correct trick is?

$\bf{\text{Edit }}\text{(Far more basic question)}:$

I get very confused with some of the measure theory notation, so I've been trying to neurotically stick to one convention so as not to be confused.

For a $\mu$-integrable $f$, I've been writing $$\int_{G}fd\mu = \int_{G}f(x)\mu(dx)$$ to mean that $x$ is my variable of integration.

Then based on this convention, if we define $\lambda(E) = \mu(yE)$, then I've been writing

$$\int_{G}fd\lambda = \int_{G}f(x)\lambda(dx) = \int_{G}f(x)\mu(y\cdot dx)$$

It this the quantity you are referring to when you write $\int_{G}f(x)d(yx)$?

If so, am I correct to perform the following calculation (returning to the case where $\mu$ is a Haar measure)?

$\begin{eqnarray*} \int_{G}fd\mu &=& \int_{G}f(x)\mu(dx)\\ &=& \int_{G}f(x)\mu(y\cdot dx)\\ &=& \int_{G}f(y^{-1}yx)\mu(y\cdot dx)\\ &=& \int_{G}f(y^{-1}x)\mu(dx)\\ &=& \int_{G}f(y^{-1}\cdot)d\mu \end{eqnarray*}$

Please be mercilessly honest if anything is even slightly incorrect. I've had a "vague" understanding of these technicalities for far too long and I want to finally really nail them down.

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To get all of these properties, use the defining property of the Haar measure: It is left-invariant under the action of $G$. In other words, $\int f(x)dx=\int f(x) d(yx)$ for any $y\in G$. Once you have this, you can change the measure in that fashion without affecting the value of the integral, and then change variables to get your old measure back: for instance, in my previous example, replace $x$ with $y^{-1}x,$ so you get $\int f(x) d(yx)=\int f(y^{-1}x) d(yy^{-1}x)=\int f(y^{-1}x)dx.$ [By the way, I'm suppressing the $\mu$ in my notation to make things look cleaner.]

Thus, we can get the first equality you ask about by replacing $y$ with $xy,$ so we get \begin{align*}(f * g)(x)&=\int_G f(y)g(y^{-1}x)dy\\&=\int_Gf(xy)g((xy)^{-1}x)d(xy)\\&=\int_Gf(xy)g(y^{-1})d(xy)\\&=\int_G f(xy) g(y^{-1})dy,\end{align*} where the final equality comes from the left-invariance of Haar measure.

The next identity you ask about can be shown in a similar fashion. For the final two, remember that the defining property of the modular function is that it reverses invariance of Haar measure: multiplying by the modular function will change your Haar measure from left-invariant to right-invariant, so those two identities will require you to use the property of right-invariance, but are otherwise the same as the above two.

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  • $\begingroup$ Thank you for your very helpful reply. I have edited my OP with a more specific (and basic) question. If my steps are correct (would you be able to confirm or correct them?), then I think I see the idea. Thanks! $\endgroup$ – roo May 12 '13 at 4:19

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