(Preamble:) I have always thought the absolutely continuity of the Lebesgue measure as: "Given a non-negative integrable function and an error bound, a sufficient reduction in the total volume of the set over which we integrate will reduce the integral of said function over the set with the reduced volume under a desired threshold".
One might also wonder the sort of a follow up to this property: Let $(X, \mathcal{B},\mu)$ be a measure space with $\mu$ the Lebsegue measure,
$$B\subset A\subset X:0<\mu(B)<\mu(A)$$
$$f\in L^1(X,\mu):\mu(\{x\in X\mid f(x) \leq 0\}) = 0$$
and define $I_A = \int_A f dx$. Now trivially $\int_{A\setminus B}f(x)dx < \int_Af(x)dx$. But what could we say about $\int_{A\setminus B}f(x)dx$ in terms of e.g. the original integral $I_A$?
(Question:) If we reduce the total volume of $A$ by some percentage, it makes intuitively (a lot of) sense that then also the value of $I_A$ is reduced by some (possibly other) percentage. But by how much? Are there any tools to classify this behavior?
If $B\subset A$, then $1_{A\setminus B}=1_A-1_B$ and therefore we have the absolute difference $$\int_A fdx-\int_{A\setminus B}fdx=\int_Bfdx$$ if all of the integrals converge properly.
Now you are asking for a relative bound, i.e. if $A'\subset A$ with $\mu(A')=p\mu(A)$ for some $p\in(0,1)$, then by the previous argument, $$\int_Afdx-\int_{A'}fdx=\int_{A\setminus A'}fdx\leq \mu(A\setminus A')\sup_{x\in A}|f(x)|$$ And with $\mu(A\setminus A')=\mu(A)-\mu(A')=\mu(A)-p\mu(A)=(1-p)\mu(A)$ you get an error bound which depends on the percentage $p$, the volume of $A$ and the magnitude of $f$.
I think it would be hard to get any meaningful bounds if we do not assume $f$ to be bounded. Consider the example where $X=(0,1)$ and $f(x)=1/x$. Then $\int_0^t fdx=\infty$ for all $t$ and you can "reduce" the interval $[0,t]$ by any percentage you'd like, but you won't get any meaningful bounds.