Statement: Let $\lambda$ be the Lebesgue measure on the real line and $f$ be a continuous function. For every $0\le\epsilon \le 1$, there is a compact set $S\subseteq [0,1]$ with $\lambda(S) = \epsilon$ and
\begin{align*} \int_0^1 f(x)\mathbf{1}_S(x)\, \lambda(\mathrm{d}x) = \epsilon\int_0^1 f(x)\,\lambda(\mathrm{d}x)\,\,. \end{align*}
Is this statement true? If so, what would be the name of this principle? I only work with discrete mathematics in my day-to-day life and I apologize if this question is too naive.
Let $c := \int_0^1f(x)\lambda(dx)$. Assume $c > 0$. Let $\bar x := x-\lfloor x\rfloor$ denote the fractional part of $x$. Note that $x \mapsto \bar x$ is measurable.
If $f$ is continuous, it is Lebesgue-measurable, therefore so is the function $\hat{f} : \mathbb{R} \to \mathbb{R}$ defined by $\hat{f}(x) := f(\bar x)$.
Since $\hat f$ is measurable, the function $F$ given by $F(s) := \int_\epsilon^{\epsilon+x}\hat{f}(x)\lambda(dx)$ is continuous.
Now we have $\int_0^1F(s)ds = \int_{0}^1\int_s^{s+\epsilon}\hat{f}(x)dxds = \int_0^\epsilon\int_0^1\hat{f}(s+x)dsdx = \int_0^\epsilon\int_0^1f(s)dsdx = \int_0^\epsilon cdx = \epsilon c$.
For this to be the case, there must be $s_0, s_2 \in [0, 1]$ such that $F(s_0) \leq \epsilon c \leq F(s_2)$. Remember that $F$ is continuous, so by the intermediate value theorem, there must also exist an $s_1$ such that $F(s_1) = \epsilon c$.
Then $S = \overline{[s_1, s_1 + \epsilon]}$ satisfies the property you are looking for.