Defining a push-forward/ image measure operator

Let $X = \mathcal{M}(\mathbb{R})$ be the space of finite signed Borel measures on $\mathbb{R}$. Given a measurable function $h\colon \mathbb{R}\rightarrow \mathbb{R}$ and $\mu \in X$, we are familiar with the image or push-forward of $\mu$ under $h$, $\mu\circ h^{-1}$ defined by:

$$\mu\circ h^{-1}(B) = \mu\big\{\omega\in \mathbb{R} \vert h(\omega)\in B\big\}= \int \textbf{1}_{B}\{h(a)\}\mu(da), \qquad B\in \mathcal{B}(\mathbb{R})$$

Where $\mathcal{B}(\mathbb{R})$ are the Borel sub-sets of $\mathbb{R}$ and $\textbf{1}_{B}\{h(a)\}=1$ if $h(a)\in B$.

Let $\lambda$ denote Lebesgue measure and let $Y= L^{1}(\mathbb{R},\lambda)$ be the normed space of equivalence classes of real-valued Lebesgue-integrable functions. My question: can we define an operator $\phi \colon Y\times X \rightarrow X$ as $h,\mu \mapsto \mu\circ h^{-1}$. Since $h$ will be an equivalence class of functions, I am concerned about what might happen when $\mu$ is the Dirac delta measure $\delta_{x}$ for $x\in \mathbb{R}$:

$$\phi(h,\delta_{x})(B) = \int \textbf{1}_{B}\{h(a)\}\delta_{x}({da}) = \textbf{1}_{B}\{h(x)\}$$

I interpret $\textbf{1}_{B}\{h(x)\}=1$ if $h^{\prime}(x)\in B$ for all $h^{\prime}$ in the equivalence class $h$. As such, I am thinking $\phi(h,\delta_{x})(B)$ cannot be a measure; consider that $\phi(h,\delta_{x})(\mathbb{R})=1$, but we have

$$\phi(h,\delta_{x})(\mathbb{R}_{++}) + \phi(h,\delta_{x})(\mathbb{R}_{-}) = 0 \not= \phi(h,\delta_{x})(\mathbb{R})=1$$

Is this reasoning valid to show that $\phi$ is not well-defined?

In order to define $L^1(\mathbb{R})$ you have to specify a specific measure $\mu\in M(\mathbb{R})$ and then we can say that $[f]$ and $[g]$ are equivalent if and only if $f=g$ $\mu$-almost everywhere. It does not make sense, when you don't specify a specific measure.
Now you are allowed to construct the mapping $$L^1(\mathbb{R},\mu) \ni [h] \stackrel{\gamma}{\mapsto} \mu \circ h^{-1} \in M(\mathbb{R})$$ This is indeed a well-defined mapping, if we can show that no one element of $L^1(\mathbb{R},\mu)$ maps to two different measures through $\gamma$. Thus fix any two $[f],[g] \in L^1(\mathbb{R},\mu)$ such that $[f]=[g]$, i.e. $f=g$ $\mu$-almost everywhere. Then we see that $$\gamma([f])(A) =\int 1_{A}(x) \, d\mu \circ f^{-1}(x)=\int 1_{A}\circ f \, d\mu = \int 1_{A} \circ g\, d\mu = \gamma([g])(A)$$ for any $A\in \mathcal{B}(\mathbb{R})$, proving that $\gamma([f])=\gamma([g])$.
• Hi Martin, sorry it was my bad. By $L^{1}(\mathbb{R})$, I meant $L(\mathbb{R},\lambda)$ where $\lambda$ is Lebesgue measure. Now if $\mu$ is not absolutely continuous with respect to $\lambda$, I do not think $h\mapsto \mu\circ h^{-1}$ is defined on $L^{1}(\mathbb{R})$ is it? – Akshay Shanker Sep 15 '16 at 2:41