Matching the integral of a function on smaller open sets

Let $$f: [0, 1] \to \mathbb R$$ be Lebesgue integrable with $$\int_0^1 f \, d \mu = C.$$

Question: For every $$K$$ with $$0 < K \leq 1$$, does there exist an open subset $$U$$ of $$[0, 1]$$ of Lebesgue measure $$K$$ such that

$$\frac{1}{K} \int_U f \, d\mu = C?$$

Remark: This is relatively easily seen to be true if $$U$$ is not required to be open. However I am not sure about the open case.

$$\newcommand\op\oplus$$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 $$\{(x,1(t\in(x,x\op 0.4)))\colon 0 for $$t=0.3$$ (left) and $$t=0.8$$ (right):