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Yes. Suppose that $f(x)$ is not a.e. constant. Then there is some subset $X$ of $\mathbb{R}$ such that $A:=f^{-1}(X)$ has positive measure but not full measure (we can take $X=(-\infty,x]$ for some adequate $x\in\mathbb{R}$).

Note that for all $d\in D$, $\mu(A\Delta(A-d))=0$. Also, as $A$ and $\mathbb{R}\setminus A$ have positive measure, by the Lebesgue density theorem there are two points $x,y$ such that, for some small value of $\varepsilon$, $\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon$ and $\mu(A\cap(y-\varepsilon,y+\varepsilon))<0.5\varepsilon$.

However, there is a sequence $(d_n)_n$ in $D$ such that $x+d_n\to y$. And for all $n$ we have $$\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))=\mu((A-d_n)\cap(x-\varepsilon,x+\varepsilon))$$ $$=\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon.$$

Thus, $\mu(A\cap(y-\varepsilon,y+\varepsilon))=\lim_{n\to\infty}\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))\geq1.5\varepsilon$, a contradiction.

Edit: It seems this argument works more generally, e.g. if $G$ is a locally compact group with $\sigma$-finite Haar measure and a measurable function $f:G\to\mathbb{R}$ satisfies that for all $d$ in some subset $D$ dense in $G$ we have $f(dx)=f(x)$ a.e., then $f$ is a.e. constant. To use the argument above in the general case one would need to use some version of the Lebesgue density theorem for locally compact groups, see e.g. Theorem A in the article "Three Results for Locally Compact Groups Connected with the Haar Measure Density", by Mueller.

Yes. Suppose that $f(x)$ is not a.e. constant. Then there is some subset $X$ of $\mathbb{R}$ such that $A:=f^{-1}(X)$ has positive measure but not full measure (we can take $X=(-\infty,x]$ for some adequate $x\in\mathbb{R}$).

Note that for all $d\in D$, $\mu(A\mathbin\Delta(A-d))=0$. Also, as $A$ and $\mathbb{R}\setminus A$ have positive measure, by the Lebesgue density theorem there are two points $x,y$ such that, for some small value of $\varepsilon$, $\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon$ and $\mu(A\cap(y-\varepsilon,y+\varepsilon))<0.5\varepsilon$.

However, there is a sequence $(d_n)_n$ in $D$ such that $x+d_n\to y$. And for all $n$ we have $$\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))=\mu((A-d_n)\cap(x-\varepsilon,x+\varepsilon))$$ $$=\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon.$$

Thus, $\mu(A\cap(y-\varepsilon,y+\varepsilon))=\lim_{n\to\infty}\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))\geq1.5\varepsilon$, a contradiction.

Edit: It seems this argument works more generally, e.g. if $G$ is a locally compact group with $\sigma$-finite Haar measure and a measurable function $f:G\to\mathbb{R}$ satisfies that for all $d$ in some subset $D$ dense in $G$ we have $f(dx)=f(x)$ a.e., then $f$ is a.e. constant. To use the argument above in the general case one would need to use some version of the Lebesgue density theorem for locally compact groups, see e.g. Theorem A in the article "Three Results for Locally Compact Groups Connected with the Haar Measure Density", by Mueller.

Yes. Suppose that $f(x)$ is not a.e. constant. Then there is some subset $X$ of $\mathbb{R}$ such that $A:=f^{-1}(X)$ has positive measure but not full measure (we can take $X=(-\infty,x]$ for some adequate $x\in\mathbb{R}$).

Note that for all $d\in D$, $\mu(A\Delta(A-d))=0$. Also, as $A$ and $\mathbb{R}\setminus A$ have positive measure, by the Lebesgue density theorem there are two points $x,y$ such that, for some small value of $\varepsilon$, $\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon$ and $\mu(A\cap(y-\varepsilon,y+\varepsilon))<0.5\varepsilon$.

However, there is a sequence $(d_n)_n$ in $D$ such that $x+d_n\to y$. And for all $n$ we have $$\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))=\mu((A-d_n)\cap(x-\varepsilon,x+\varepsilon))$$ $$=\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon.$$

Thus, $\mu(A\cap(y-\varepsilon,y+\varepsilon))=\lim_{n\to\infty}\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))\geq1.5\varepsilon$, a contradiction.

Yes. Suppose that $f(x)$ is not a.e. constant. Then there is some subset $X$ of $\mathbb{R}$ such that $A:=f^{-1}(X)$ has positive measure but not full measure (we can take $X=(-\infty,x]$ for some adequate $x\in\mathbb{R}$).

Note that for all $d\in D$, $\mu(A\Delta(A-d))=0$. Also, as $A$ and $\mathbb{R}\setminus A$ have positive measure, by the Lebesgue density theorem there are two points $x,y$ such that, for some small value of $\varepsilon$, $\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon$ and $\mu(A\cap(y-\varepsilon,y+\varepsilon))<0.5\varepsilon$.

However, there is a sequence $(d_n)_n$ in $D$ such that $x+d_n\to y$. And for all $n$ we have $$\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))=\mu((A-d_n)\cap(x-\varepsilon,x+\varepsilon))$$ $$=\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon.$$

Thus, $\mu(A\cap(y-\varepsilon,y+\varepsilon))=\lim_{n\to\infty}\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))\geq1.5\varepsilon$, a contradiction.

Edit: It seems this argument works more generally, e.g. if $G$ is a locally compact group with $\sigma$-finite Haar measure and a measurable function $f:G\to\mathbb{R}$ satisfies that for all $d$ in some subset $D$ dense in $G$ we have $f(dx)=f(x)$ a.e., then $f$ is a.e. constant. To use the argument above in the general case one would need to use some version of the Lebesgue density theorem for locally compact groups, see e.g. Theorem A in the article "Three Results for Locally Compact Groups Connected with the Haar Measure Density", by Mueller.

Yes. Suppose that $f(x)$ is not a.e. constant. Then there is some subset $X$ of $\mathbb{R}$ such that $A:=f^{-1}(X)$ has positive measure but not full measure (we can take $X=(-\infty,x]$ for some adequate $x\in\mathbb{R}$).

Note that for all $d\in D$, $\mu(A\Delta(A-d))=0$. Also, as $A$ and $\mathbb{R}\setminus A$ have positive measure, by the Lebesgue density theorem there are two points $x,y$ such that, for some small value of $\varepsilon$, $\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon$ and $\mu(A\cap(y-\varepsilon,y+\varepsilon))<0.5\varepsilon$.

However, there is a sequence $(d_n)_n$ in $D$ such that $x+d_n\to y$. And for all $n$ we have $$\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))=\mu((A-d_n)\cap(x-\varepsilon,x+\varepsilon))=\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon.$$

Thus, $\mu(A\cap(y-\varepsilon,y+\varepsilon))=\lim_{n\to\infty}\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))\geq1.5\varepsilon$, a contradiction.

Yes. Suppose that $f(x)$ is not a.e. constant. Then there is some subset $X$ of $\mathbb{R}$ such that $A:=f^{-1}(X)$ has positive measure but not full measure (we can take $X=(-\infty,x]$ for some adequate $x\in\mathbb{R}$).

Note that for all $d\in D$, $\mu(A\Delta(A-d))=0$. Also, as $A$ and $\mathbb{R}\setminus A$ have positive measure, by the Lebesgue density theorem there are two points $x,y$ such that, for some small value of $\varepsilon$, $\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon$ and $\mu(A\cap(y-\varepsilon,y+\varepsilon))<0.5\varepsilon$.

However, there is a sequence $(d_n)_n$ in $D$ such that $x+d_n\to y$. And for all $n$ we have $$\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))=\mu((A-d_n)\cap(x-\varepsilon,x+\varepsilon))$$ $$=\mu(A\cap(x-\varepsilon,x+\varepsilon))>1.5\varepsilon.$$

Thus, $\mu(A\cap(y-\varepsilon,y+\varepsilon))=\lim_{n\to\infty}\mu(A\cap(x+d_n-\varepsilon,x+d_n+\varepsilon))\geq1.5\varepsilon$, a contradiction.