Verifying Convolution Identities 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.
 A: 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.
