understanding the definition of measurable mappings In definition of Measurable mappings, we consider measurable spaces $(\Omega_1, \mathcal{F}_1)$ and $(\Omega_2, \mathcal{F}_2)$ and the mapping $T:\Omega_1 \rightarrow \Omega_2$ is measurable $\mathcal{F}_1/\mathcal{F}_2$ if $T^{-1}A \in \mathcal{F}_1$ for each $A\in \mathcal{F}_2$.
my question is why we cannot define like '$TA\in\mathcal{F}_2$ for all $A\in\mathcal{F}_1$'. What is the fallacy in this definition? 
Thanks in advance
 A: Consider a simple example, where $T:(\mathbb{R},\mathcal{B}(\mathbb{R}))\to (\mathbb{R},\{\emptyset,\Omega\})$ is given by $T(x)=x$. It is clear that $f$ is measurable according to the "standard" definition. However, $T([0,1])\notin \{\emptyset,\Omega\}$. In fact, a function that satisfies both conditions is called bimeasurable.
So, what is the rationale behind the "standard" definition? It is useful for developing of the integration theory because measurable functions (in the "standard" sense) exhibit a lot of appealing properties. For example, the space of Borel-measurable (real-valued) functions is a vector space which is closed under pointwise limits.
A: 
As I mentioned in the comments measurability of functions is defined precisely so as to render all continuous functions (with respect to the respective Borel algebras) measurable. So why is that?

First note, that if $(X, T_X)$ and $(Y, T_Y)$ are topological spaces, then by definition a continuous function $f \colon X \to Y$ satisfies $$\forall A \in T_Y \colon f^{-1}(A) \in T_X$$ (This would give the person trying to think up a proper definition for measurable functions a first incentive)
Recall, that if you have some set $Z$ and some set of subsets $\mathcal{S} \subset \mathcal{P}(Z)$, where $\mathcal{P}(Z) = \{A \mid A \subset Z\}$ is the power set of $Z$, then the smallest $\sigma$-algebra generated by $\mathcal{S}$ is denoted by $\sigma(\mathcal{S})$. In other words $$\sigma(\mathcal{S}) = \bigcap\limits_{\mathcal{S} \subset \mathcal{A} \text{, } \mathcal{A} \text{ is a } \sigma-\text{algebra}} \mathcal{A}$$
Now think back to the topological spaces $(X, T_X)$ and $(Y, T_Y)$. The Borel sigma algebra with respect to $(X, T_X)$, denoted by $B(X)$, is defined to be the smallest $\sigma$-algebra that contains all open subsets of $X$, i.e. $B(X) = \sigma(T_X)$. Analogously, $B(Y) = \sigma(T_Y)$.
Now we stick to the convention, that if $f \colon L' \to L$ is any function, with $L, L'$ some sets, and $\mathcal{E} \subset \mathcal{P}(L)$, then we define $$f^{-1}[\mathcal{E}] = \{f^{-1}(E) \mid E \in \mathcal{E}\} \subset \mathcal{P}(L')$$
Having dealt with these definitions, we prove a rather interesting fact:


Theorem: Let $L, L'$ be some sets and $\mathcal{E} \subset \mathcal{P}(L)$. If $f \colon L' \to L$ is some function, then $$f^{-1}[\sigma(\mathcal{E})] = \sigma(f^{-1}[\mathcal{E}])$$

Proof: It is not hard to show, that $f^{-1}[\sigma(\mathcal{E})]$ is a $\sigma$-algebra. Now we have $$f^{-1}[\mathcal{E}] \subset f^{-1}[\sigma(\mathcal{E})] \implies \sigma(f^{-1}[\mathcal{E}]) \subset \sigma(f^{-1}[\sigma(\mathcal{E})]) = f^{-1}[\sigma(\mathcal{E})]$$
Thus it remains to prove, that $f^{-1}[\sigma(\mathcal{E})] \subset \sigma(f^{-1}[\mathcal{E}])$.
For this we use the principle of good sets. So define $$\mathcal{G} = \big\{A \in \sigma(\mathcal{E}) \mid f^{-1}(A) \in \sigma(f^{-1}[\mathcal{E}]) \big\}$$ Our claim holds, if we can prove $\mathcal{G} = \sigma(\mathcal{E})$. But this is actually not too hard, since clearly $\mathcal{E} \subset \mathcal{G}$, so we only need to show, that $\mathcal{G}$ is a $\sigma$-algebra. Clearly since $f^{-1}(L) = L' \in \sigma(f^{-1}[\mathcal{E}])$ we have $L \in \mathcal{G}$. Now if $A \in \mathcal{G}$, then $$f^{-1}(A) \in \sigma(f^{-1}[\mathcal{E}]) \implies f^{-1}(A)^c = f^{-1}(A^c) \in \sigma(f^{-1}[\mathcal{E}]) \implies A^c \in \mathcal{G}$$ Last but not least, if $\{A_n\}_{n \in \mathbb{N}} \subset \mathcal{G}$, then $$f^{-1}\bigg(\bigcup\limits_{n \in \mathbb{N}} A_n \bigg) = \bigcup\limits_{n \in \mathbb{N}} \underbrace{f^{-1}(A_n)}_{\in \sigma(f^{-1}[\mathcal{E}])} \in \sigma(f^{-1}[\mathcal{E}])$$ which shows that $\mathcal{G}$ is a $\sigma$-algebra and completes our proof.



Now why was that interesting?

Corollary: If $(L, \Sigma_L)$ and $(L', \Sigma_{L'})$ are measure spaces and $l \subset \mathcal{P}(L)$ is a generator for $\Sigma_L$ i.e. $\sigma(l) = \Sigma_L$ and if $f \colon L' \to L$ is a function, then $f$ is measurable (with respect to these measure spaces) if and only if $f^{-1}[l] \subset \Sigma_{L'}$.

Proof: It is clear, that if $f$ is measurable we have $f^{-1}[l] \subset \Sigma_L$. So suppose now, that $f^{-1}[l] \subset \Sigma_L$, then by the preceding theorem we have $$f^{-1}[\Sigma_L] = f^{-1}[\sigma(l)] = \sigma(f^{-1}[l]) \subset \Sigma_{L'}$$ So $f$ is indeed measurable.


The content of this corollary tells us, that measurability of functions is a property that only needs to be verified on generators of the sigma-algebras in question (another analogy to topology - recall sub-bases for topologies). Now at last we go back to our topological spaces $(X, T_X)$ and $(Y, T_Y)$ with some continuous function $f: X \to Y$. Consider the measure spaces $(X, B(X))$ and $(Y, B(Y))$. By the corollary, in order to show measurability of our function $f$, it suffices to show $$f^{-1}[T_Y] \subset B(X)$$ But this is satisfied by definition of a continuous function, since $f^{-1}[T_Y] \subset T_X \subset B(X)$, which explains what I meant in the comments.
