Show that $\int_0^1\frac{1}{|f(x)-x_0|}dx$ is unbounded

Question

Let $f$ be a Lebesgue measurable functinon on such that $f:[0,1]\rightarrow [0,1]$. Prove that for any $M$ there exists $x_0\in[0,1]$ such that $$ \int_0^1\frac{1}{|f(x)-x_0|}dx\geq M. $$

My attempt:

There exists a sequence of step functions $f_n$ such that $f_n$ increasing to $f$ pointwise and $$ \int_0^1f_n(x)dx=\int_0^1f(x)dx. $$

I am not sure whether we can prove that $$ \lim\int_0^1\frac{1}{|f_n(x)-x_0|}dx=\int_0^1\frac{1}{|f(x)-x_0|}dx $$ since the denominator is not bounded. If above is true, then we can get the desired result.

Answer 1

Prove by contradiction. If the integral is bounded by some $M$ for all $x_0$ then $\int_0^{1}\int_0^{1} \frac 1 {|f(x)-y|} dxdy \leq M <\infty$. By Fubini/Tonelli Theorem we can write this as $\int_0^{1}\int_0^{1} \frac 1 {|f(x)-y|} dydx \leq M$ This is a contradiction because the inside integral is $\infty$ for every $x$.