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Why do all measure theory textbooks present the concept of push-forward measure, but never the concept of pull-back measure? Doesn't the latter exist?

It's true that the naive treatment of such a concept would sometimes lead to contradictions. For instance, let $p:\mathbb{R}^2\rightarrow\mathbb{R}$ be given by $p(x,y)=x$, $\lambda$ be the Lebesgue measure on $\mathbb{R}$, and $A=[0,1]\times[0,1]$. If one decomposes $A=A_1 \cup A_2$ where $A_1=[0,1]\times[0,\frac{1}{2}]$ and $A_2=[0,1]\times(\frac{1}{2},1]$, then $(p^* \lambda)(A)=1$ while $(p^* \lambda)(A_1)+(p^* \lambda)(A_2)=2$, showing that the naive definition of the pull-back does not lead to a measure.

Similarily, if $i:\mathbb{R}\rightarrow\mathbb{R}^2$ is given by $i(x)=(x,0)$, then the pull-back of the Lebesgue measure on $\mathbb{R}^2$ would be 0.

Given the two situations presented above, can one define a fruitful concept of pull-back measure?

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# Pullback measures

Why do all measure theory textbooks present the concept of push-forward measure, but never the concept of pull-back measure? Doesn't the latter exist?

It's true that the naive treatment of such a concept would sometimes lead to contradictions. For instance, let $p:\mathbb{R}^2\rightarrow\mathbb{R}$ be given by $p(x,y)=x$, $\lambda$ be the Lebesgue measure on $\mathbb{R}$, and $A=[0,1]\times[0,1]$. If one decomposes $A=A_1 \cup A_2$ where $A_1=[0,1]\times[0,\frac{1}{2}]$ and $A_2=[0,1]\times(\frac{1}{2},1]$, then $(p^* \lambda)(A)=1$ while $(p^* \lambda)(A_1)+(p^* \lambda)(A_2)=2$, showing that the naive definition of the pull-back does not lead to a measure.

Similarily, if $i:\mathbb{R}\rightarrow\mathbb{R}^2$ is given by $i(x)=(x,0)$, then the pull-back of the Lebesgue measure on $\mathbb{R}^2$ would be 0.

Given the two situations presented above, can one define a fruitful concept of pull-back measure?