Definitions:
Let $\lambda$ be the Lebesgue measure on $G$. Define Haar measure $\mu$ on $G$ by $\mathrm{d}\mu(x) := \frac{1}{x}\mathrm{d}\lambda(x)$.
$G$ is multiplicative group, $G=(R_+,⋅)$. Since my functions are $L^1(\mathbb{R})$, they are well behaved on $[1,\infty)$, so I am interested in $[0,1]$, where the $1/x$ in $\mu$ might blow up the finiteness of measure. Here, suppose $G = [0,1]$.
Motivation:
It is simple to show $\|f\|_{L^p(G,\lambda)} \leq \| f \|_{L^p(G,\mu)}$. Thus,
\begin{equation} L^p(G,\mu) \subset L^p(G,\lambda). \end{equation}
I am trying to understand the difference between $L^{\infty}(G,\lambda)$ versus $L^{\infty}(G,\mu)$.
Question 1: I like to take the limit $p \to \infty$ to obtain $L^{\infty}(G,\mu) \subset L^{\infty}(G,\lambda)$. Is there any issue with this? I know that
\begin{equation} \lambda(G) < \infty \quad \Rightarrow \quad \lim_{p\to \infty} \| f \|_{L^p(G,\lambda)} = \| f \|_{L^{\infty}(G,\lambda)} \end{equation}
On the other hand, $\mu(G) = \infty$, so I am not sure taking the limit $p \to \infty$ is still valid. However, $\mu$ is $\sigma$-finite. Can I still conclude the following?
\begin{equation} \lim_{p\to \infty} \| f \|_{L^p(G,\mu)} = \| f \|_{L^{\infty}(G,\mu)} \end{equation}
Question 2: Do all bounded functions on $G$ are also in $L^{\infty}(G,\mu)$? I guess yes, because, by definition
\begin{equation} L^{\infty}(G,\mu) = \inf_a \{ \mu(\{x: f(x) > a\}) = 0 \} \end{equation} When $|f| < M < \infty$ for all $x \in G$, then $\{x: f(x) > M\} = \emptyset$, and $\mu(\emptyset) = 0$ thus $\| f \|_{L^{\infty}(G,\mu)} \leq M$.
Question 3: Do all essentially bounded (with respect to $\lambda$) functions on $G$ are in $L^{\infty}(G,\mu)$? In effect, this would mean
\begin{equation} L^{\infty}(G,\lambda) \subset L^{\infty}(G,\mu), \end{equation} and combined with $L^{\infty}(G,\mu) \subset L^{\infty}(G,\lambda)$ (in question 1), yields $L^{\infty}(G,\lambda) = L^{\infty}(G,\mu)$. Is this true?
To show question 3, I should know if zero measure by $\lambda$ implies zero measure by $\mu$. In other words, does
\begin{equation} \mu \ll \lambda \end{equation} hold ($\mu$ is absolutely continuous)?
I am learning measure theory very recently, so please elaborate your answers if possible. Thank you.
For Q 1: Yes, $\|f\|_p \to \|f\|_{\infty}$ as $ p \to \infty$ even for infinite measures.
For Q 2: Yes, bounded functions belong to $L^{\infty}(\mu)$ even for infinite measures $\mu$.
For Q 3: $\mu(E)=\int_E \frac 1 x d\lambda(x)=0$ whenever $\lambda (E)=0$ so $\mu << \lambda$ so $L^{\infty}(\lambda) \subset L^{\infty}(\mu)$.