Is there measurable function f: $\mathrm\forall a\int_{0}^{1}{\frac{1}{f(x)-a}}\, d\mu<\infty$?

Question

I would like to find measurable function f: $\mathrm\forall a\int_{0}^{1}{\frac{1}{f(x)-a}}\, d\mu<\infty$, but I can't find any such function. I know, that f can't be continuous. If f is continuous, we can prove that such integral is not finite by contradiction, using Fatou and Fubini's theorem. However I can't prove there is no such function, if f can be non-continious. Could anybody help me?

Answer 1

For simplicity we can assume that $f^{-1}(c)$ has measure zero for any $c$ (otherwise take $a = c$ and integral isn't even defined).

There is some segment $[x, y]$ s.t. $\mu(f^{-1}([x, y])) > \varepsilon$. W.l.o.g. assume $x = 0$, $y = 1$.

If we take $a \in [0, 1]$ we have $\int_0^1 \frac{1}{|f(x) - a|}\, d\mu > \int_{f^{-1}([0, 1])}\frac{1}{|f(x) - a|}\, d\mu > \varepsilon$.

Now, either $\mu(f^{-1}([0, \frac{1}{2}])) \geqslant \frac{\varepsilon}{2}$ or $\mu(f^{-1}([\frac{1}{2}, 1])) \geqslant \frac{\varepsilon}{2}$. Assuming the former, we get for $a \in [0, \frac{1}{2}]$, $\int_0^1 \frac{1}{|f(x) - a|}\, d\mu > \frac{\varepsilon}{2} + \varepsilon$ (integral over some subset of $f^{-1}([0, \frac{1}{2}])$ with measure $\frac{\varepsilon}{2}$ is at least $\varepsilon$, and integral over the rest of $f^{-1}([0, 1])$ is at least $\frac{\varepsilon}{2}$).

Splitting $[0, \frac{1}{2}]$ further, we get that for example $\mu(f^{-1}([0, \frac{1}{4}])) > \frac{\varepsilon}{4}$, and if $a \in [0, \frac{1}{4}]$, then $\int_0^1 \frac{1}{|f(x) - a|}\, d\mu > \frac{\varepsilon}{2} + \frac{\varepsilon}{2} + \varepsilon$.

We continue doing so, and get sequence of nested segments $X_n$ with length $\frac{1}{2^n}$ s.t. if $a \in X_n$ then $\int_0^1\frac{1}{|f(x) - a|}\, d\mu > n \cdot \frac{\varepsilon}{2} + \varepsilon$. Taking their intersection, we find value of $a$ s.t. integral diverges (as it's greater than $n \cdot \frac{\varepsilon}{2}$ for any $n$).