$\newcommand\op\oplus$Apparently, here $\mu$ is the Lebesgue measure. Identify the interval $[0,1)$ with the (say) unit circle in a standard manner. Slightly more elementarily, for $x$ and $K$ in $(0,1)$, let 
$$
(x,x\op K):=
\begin{cases}
(x,x+K)&\text{ if }x+K\le1,\\
(x,1)\cup(0,x+K-1)&\text{ if }x+K>1. 
\end{cases}
$$ 
Then 
$$\int_0^1 dx\,\int_{(x,x\op K)} dt\,f(t)
=\int_0^1 dt\,f(t)\,\int_0^1 dx\,1(t\in(x,x\op K))
=\int_0^1 dt\,f(t)\,K.$$
Since $\int_{(x,x\op K)} dt\,f(t)$ is continuous in $x$, there exists some $x_*\in[0,1]$ such that 
$$\int_{(x_*,x_*\op K)} dt\,f(t)=\int_0^1 dt\,f(t)\,K,$$
as desired. 

---

Here are the graphs $\{(x,1(t\in(x,x\op 0.4)))\colon 0<x<1\}$ for $t=0.3$ (left) and $t=0.8$ (right): 

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/yVwW6.png