Does there exist a measurable function with a specific property? Does there exist such a measurable function $f$ from $[0 ; 1]$ to $\mathbb{R}$ (under Lebesgue measure) such that for  $\forall x \in \mathbb{R}$ $\int_0^1 \frac{dt}{|f(t) - x|}$ is always finite?
(The integral in this question is a Lebesgue integral)
I failed to construct any examples, but I have not found any way to prove that such functions do not exist either. 
Any help will be appreciated.
 A: It's not possible.  Let $M$ be such that the set of $t$ with $|f(t)|<M$ has measure at least $1/2$. Then divide the interval $I_0=[-M,M]$ into halves at it midpoint, and choose the half $I_1$ with $\mu(f^{-1}(I_1))\ge\mu(f^{-1}(I_0))/2$.  Keep splitting the interval in half at the midpoint and choosing the half whose inverse image has at least half the measure of the inverse image of the previous one.  These intervals converge to a point $x$.  You will see that $|f(t)-x|$ is less than $M/2^{n-1}$ on a set $f^{-1}(I_n)$ of measure at least $(1/2)^{n+1}$.  Now, starting at any $k$ and counting downward, you have $\frac{1}{|f(t)-x|}$ greater than or equal to $2^{k-1}/M$ on a set of measure at least $(1/2)^{k+1}$.  Furthermore, it is greater than or equal to $2^{k-2}/M$ on a set of measure at least $(1/2)^k$, ...., and at least $2^{-1}/M$ on a set of measure at least $1/2$.  You can see that this will imply the integral is greater than or equal to $$\frac{2^{k-1}}{M}\cdot\frac{1}{2^{k+1}} + \frac{2^{k-2}}{M}\cdot(\frac{1}{2^{k}} - \frac{1}{2^{k+1}}) + \frac{2^{k-3}}{M}\cdot(\frac{1}{2^{k-1}} - \frac{1}{2^{k}}) + \cdot\cdot\cdot + \frac{2^{-1}}{M}\cdot(\frac{1}{2^{1}} - \frac{1}{2^{2}}) $$, which simplifies to
$$\frac{2^{k-1}}{M}\cdot\frac{1}{2^{k+1}} + \frac{2^{k-2}}{M}\cdot\frac{1}{2^{k+1}} + \frac{2^{k-3}}{M}\cdot\frac{1}{2^{k}} + \cdot\cdot\cdot + \frac{2^{-1}}{M}\cdot\frac{1}{2^{2}} $$, and then to $$\frac{1}{4M}+k\cdot\frac{1}{8M}$$, which is more than $\frac{k}{8M}$.
Note that the integral is bounded below by this sum.  Since for some fixed $M$ this holds for arbitrarily large $k$, this lower bound can be made as large as you like, so for this particular $x$ the interval must diverge.
