Sufficient condition for measurability

Question

Let $f$ be a real-valued function defined on a measurable set $E$. Show that the measurability of set {$x$ : $f(x)$ = c} , where c $\in$ $\Bbb R$, is not sufficient for $f$ to be measurable.

Answer 1

Let $E=[0,1]$ and let $\mathcal{A}$ be the sigma-algebra of the subsets of $E$ that are either countable or their complement in $[0,1]$ is countable. Take the function $f:E\to\mathbb{R}$ to be $f(x)=x$. Then $f$ is non-measurable: Indeed, if $c\in(0,1)$ then $f^{-1}(-\infty,c)=[0,c)$ which is not measurable (not an element of $\mathcal{A}$). However, $\{x\in E: f(x)=c\}=\{c\}$ when $c\in[0,1]$ and $\{x\in E: f(x)=c\}=\emptyset$ when $c\not\in [0,1]$ which is measurable in any case (because $\mathcal{A}$ contains singletons and the empty set).

Edit

A second example, just in case OP refers to Lebesgue measurability only.

Take $E=\mathbb{R}$ and let $N$ be a non-measuralbe subset of $[0,1]$. Such subsets exist, due to the axiom of choice. Now take $f(x)=x+\chi_N(x)$, where $\chi_N$ is the characteristic function of $x$, i.e. $\chi_N(x)=1$ when $x\in N$ and $\chi_N(x)=0$ when $x\not\in N$. The function $\chi_N$ is a non-measurable function obviously. Now $f$ is non-measurable: the identity function $x\mapsto x$ is of course Lebesgue measurable and it is very easy to prove that the sum of a measurable and a non-measurable function is non-measurable.

On the other hand, if $c\in\mathbb{R}$ then $\{x\in\mathbb{R}: f(x)=c\}=\{x\in N: x+1=c\}\cup\{x\in \mathbb{R}\setminus N: x=c\}$. Now we distinguish cases:

Case $1$: if $c\in \mathbb{R}\setminus N$ and $c-1\in N$, then $\{x\in \mathbb{R}: f(x)=c\}=\{c\}\cup\{c-1\}$.

Case $2$: if $c\in \mathbb{R}\setminus N$ and $c-1\in \mathbb{R}\setminus N$ then $\{x\in \mathbb{R}: f(x)=c\}=\{c\}$

Case $3$: if $c\in N$ and $c-1\in N$ then $\{x\in \mathbb{R}: f(x)=c\}=\{c-1\}$.

Case $4$ if $c\in N$ and $c-1\in\mathbb{R}\setminus N$ then $\{x\in \mathbb{R}: f(x)=c\}=\emptyset$.

So in any case, i.e. for any $c\in\mathbb{R}$, this is a Lebesgue-measurable set.

Answer 2

Let $S$ be a non-measurable subset of $\mathbb{R}$ and let $S^c=\mathbb{R}{\setminus}S$.

Let $g$ be an injective function from $S$ into $(0,1)$, and let $h$ be an injective function from $S^c$ into $(-1,0)$.

Now define $f:\mathbb{R}\to\mathbb{R}$ by $$ f(x) = \begin{cases} g(x)&\text{if}\;x\in S\\[4pt] h(x)&\text{if}\;x\not\in S\\ \end{cases} $$ Then $f$ is injective, hence the inverse image of any singleton is measurable (since it has cardinality at most one).

But $f$ is not measurable since the inverse image of $(0,1)$ is $S$.