$\newcommand\op\oplus$Apparently, here $\mu$ is the Lebesgue measure. Let $\op$ denote the addition $\mod 1$, so that $x\op y=x+y-1$ for $x,y$ in $[0,1)$ such that $x+y\ge1$. Then 
$$\int_0^1 dx\,\int_{(x,x\op K)} dt\,f(t)
=\int_0^1 dt\,f(t)\,\int dx\,1(t\in(x,x\op K)
=\int_0^1 dt\,f(t)\,K.$$
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.