Haar measure for large locally compact groups In this answer, Gerald Edgar mentions that Haar measure is naturally defined on the $\sigma$-algebra of Baire sets (the smallest $\sigma$-algebra that contains all the compact $G_\delta$ sets) of a locally compact group and that the uniqueness of Haar measure can fail for larger $\sigma$-algebras. I wonder if there is a nice example of this.
Curious, I skimmed through Halmos's classic Measure Theory, and I found that he proves the existence and uniqueness of Haar measure for the slightly larger $\sigma$-algebra generated by all compact sets. (Confusingly, Halmos defines Borel sets to be those in this $\sigma$-algebra; I will stick with the usual definition of Borel sets.)

Is there a nice example of a locally compact group where the uniqueness of Haar measure fails for the $\sigma$-algebra of Borel sets — the $\sigma$-algebra generated by open sets?

To dispell some potential confusion (see comments by Keenan Kidwell and Gerald Edgar) Haar measures are not required to be regular (for the purpose of this question).
 A: I may be missing something here but what François was looking for may be the following (it is of course already present in the previous answers, I only put it slightly differently):
Let $B$ be a Borel subset of the group $\mathbb{R} \times \mathbb{R}_d$. Define $B_y = \{x : (x,y) \in B\}$. Denote the Lebesge measure on $\mathbb{R}$ by $\lambda$.
First, let $$\mu(B) = \sum_{y \in \mathbb{R}} \lambda(B_y).$$ Second, let 
$$
\nu(B) = 
\begin{cases}
\sum_{y \in \mathbb{R}} \lambda(B_y) \textrm{ if } B \textrm{ only intersects countably many sets of the form } \mathbb{R} \times \{y\},\\
\infty, \textrm{ otherwise. }
\end{cases}
$$
It is easy to check that both $\mu$ and $\nu$ are translation invariant $\sigma$-additive measures on the Borel sets, taking finite values on the compact sets and positive values on the nonempty open sets. (And the open sets are inner regular w.r.t. compact sets.) Do you require anything else? 
Moreover, $\mu(\{0\} \times \mathbb{R}) = 0$, but $\nu(\{0\} \times \mathbb{R}) = \infty$.
A: Let's try this example.  Let $\mathbb R$ be the additive group of reals with the usual topology, and let $\mathbb R_\mathrm{d}$ be the additive group of reals with the discrete topology.  Our group is $G = \mathbb R \times \mathbb R_\mathrm{d}$.  A big group.  Any compact set in $G$ has finite projection on the $y$-axis.  The Haar integral $\Lambda$, defined on $C_c(G)$, is
$$
\Lambda(f) = \sum_y \int f(x,y)\,dx
$$
where of course the sum is finite.  
Now consider the (Borel but not Baire) set $E = \{0\}\times \mathbb R_\mathrm{d}$.  It is closed, hence Borel.  But any Baire set has projection on the $y$-axis either countable or co-countable, so $E$ is not a Baire set.  The set $E$ has outer measure $\infty$ but inner measure $0$.  Changing our "extension" of Haar measure on $E$ to any value between $0$ and $\infty$ will make sense, and still be a Borel measure that produces $\Lambda$ on the functions in $C_c(G)$.
