Let $f:[0,1] \to [0,1]$ be a Lebesgue integrable function on $[0,1]$. Prove that there exists $t \in [0,1]$ such that $\frac{1}{|t-f(x)|}$ is not Lebesgue integrable on $[0,1]$.
This is an interesting problem but I have no idea at all. Any hints would be appreciated.
Edit: Here are some of my attempts. First I noticed that if $\{x:f(x)=t\}$ is of positive measure, then clearly it is done. While this may not hapen, I tried to consider the set $\{x:|f(x)-t|<c\}$ instead. I also tried Chebyshev's inequality. But these seem to be of no use.
Here is one way that goes back to basics: For each positive integer $k$, divide the interval $[0,1]$ (which is the target set of $f$) into $2^k$ equal length disjoint intervals. Let $$I_i(k)=[a_i(k),b_i(k)) \quad \forall i \in \{1, ..., 2^k\}$$ denote the particular intervals (the last interval has the form $[a_{2^k}(k), b_{2^k}(k)]$). Now:
For $k=1$ we have $I_1(1)=[0, 1/2), I_2(1)=[1/2,1]$. Let $i^*(1) \in \{1,2\}$ denote the index $i$ that satisfies $$\mu\{x \in [0,1]:f(x)\in I_{i}\} \geq 1/2$$ (if there are ties, choose the smallest such $i$). Let $I^*(1)$ denote the corresonding interval $I_i(k)$ for $i=i^*(k)$.
For $k=2$, chop interval $I^*(1)$ in two and choose the half that has at least half the probability, call that $i^*(2)\in\{1, ..., 4\}$ and let $I^*(2)$ denote the corresponding interval. Then $$ \mu\{x\in [0,1]: f(x) \in I^*(2)\} \geq 1/4$$
We proceed with nested intervals $$ I^*(1)\supseteq I^*(2)\supseteq I^*(3)...$$ such that for all $k$ we have $|I^*(k)|=1/2^k$ and $$ \mu\{x\in[0,1]:f(x) \in I^*(k)\} \geq 1/2^k \quad (Eq. *)$$ We now use the nested interval theorem, which (unfortunately) requires closed intervals, so let us take the closure $\overline{I}^*(k)$. The nested interval theorem implies there exists a point $t \in \cap_{k=1}^{\infty} \overline{I}^*(k)$.
As a minor cosmetic issue, we want to say $t$ is in all the intervals $I^*(k)$, but we can only say it is in all intervals $\overline{I}^*(k)$. So we redefine for each $k \in \{1, 2, 3, ...\}$ and each $i\in\{1, ..., 2^k\}$ new intervals $D_i(k)$ that again partition $[0,1]$ into disjoint intervals of length $1/2^k$, such that for each $i \in \{1, ..., 2^k\}$ we have $$ D_{i}(k) = \left\{\begin{array}{cc} I_i(k) & \mbox{ if $i \notin \{i^*(k), i^*(k)+1\}$} \\ \overline{I}^*(k) & \mbox{ if $i=i^*(k)$}\\ (a_i(k),b_i(k)) & \mbox{ if $i=i^*(k)+1$} \end{array}\right. $$ and call $D^*(k)=I^*(k)$. This simply ensures:
Property 1: $t \in D^*(k)$ for all $k \in \{1, 2, 3, ...\}$.
Property 2: $\mu\{x \in [0,1]: f(x)\in D^*(k)\}\geq 1/2^k$ for all $k \in \{1, 2, 3, ...\}$.
Note that Property 2 follows from (Eq. *) because $I^*(k)\subseteq D^*(k)$.
Now assume $$ \int \frac{1}{|t-f(x)|}dx<\infty$$ (we reach a contradiction). Define $S=\{x \in [0,1]: f(x)\neq t\}$. and observe that $\mu\{S\}=1$. Define $h:[0,1]\rightarrow[0,\infty)$ by $$ h(x) = \left\{\begin{array}{cc} \frac{1}{|t-f(x)|} & \mbox{if $x \in S$} \\ 0 & \mbox{ else} \end{array}\right.$$ Since $h(x)=1/|t-f(x)|$ for almost all $x \in [0,1]$, we have $$\int_0^1 h(x)dx = \int_0^1 \frac{1}{|t-f(x)|}dx$$ Let us construct a simple function: For each $k\in \{1, 2, 3, ...\}$ and each $i \in \{1, ..., 2^k\}$ define $$v_i(k) = \inf_{y \in D_i(k)} \frac{1}{|t-y|}$$ Observe that $$v_{i^*(k)}(k)\geq 2^{k} \quad \forall k \in \{1, 2, 3, ...\} \quad (Eq. **)$$ Then define $$g_k(x) = \sum_{i=1}^{2^k}v_i(k)1_{\{x \in D_i(k)\cap S\}}\quad \forall x \in [0,1]$$ $$ \tilde{g}_k(x) = \sum_{i \in \{1, ..., 2^k\}\setminus i^*(k)}v_i(k)1_{\{x \in D_i(k)\cap S\}}\quad \forall x \in [0,1]$$ We get $$ 0\leq \tilde{g}_k(x)\leq g_k(x)\leq h(x)\quad \forall x \in S$$ Further $$ \tilde{g}_k\nearrow h $$ so $$ \sum_{i \in \{1, ..., 2^j\}\setminus i^*(k)}v_i(k)\mu\{x\in [0,1]:f(x)\in D_i(k)\} \nearrow \int h$$ However, since $\tilde{g}_k\leq g_k\leq h$ we have $\int \tilde{g}_k\leq \int g_k\leq \int h$ and in particular, since $\int h<\infty$, we have $\lim_{k\rightarrow\infty}\left[(\int g_k)-(\int \tilde{g}_k)\right] = 0$. So $$ \lim_{k\rightarrow\infty}v_{i^*(k)}(k)\mu\{x \in [0,1]:f(x)\in D^*(k)\}= 0$$ On the other hand we know for all $k\in\{1, 2, 3, ...\}$ that $$ v_{i^*(k)}(k)\mu\{x \in [0,1]:f(x)\in D^*(k)\} \geq 2^k\frac{1}{2^k}=1$$ where we have used Property 2 and (Eq **). This is a contradiction. $\Box$