Suppose that $ f:[a,b] \rightarrow \mathbb{R} $ is continuous on $(a, b)$ and has singularities at $a$ and $b$.
Let $c \in (a, b)$.
I think the following statements are true, but I cannot prove (1).
(1)
If $$\lim_{h \to 0+, h' \to 0+} \int_{a+h}^{b-h'} f = A$$ for some real number $A$, then $$\lim_{h \to 0+} \int_{a+h}^{c} f = B$$ for some real number $B$ and $$\lim_{h \to 0+} \int_{c}^{b-h} f = C$$ for some real number $C$ and $$A = B + C.$$
(2)
If $$\lim_{h \to 0+} \int_{a+h}^{c} f = B$$ for some real number $B$ and $$\lim_{h \to 0+} \int_{c}^{b-h} f = C$$ for some real number $C$, then $$\lim_{h \to 0+, h' \to 0+} \int_{a+h}^{b-h'} f = A$$ for some real number $A$ and $$A = B + C.$$
Proof of (2):
$$\lim_{h \to 0+} \int_{a+h}^{c} f = B.$$
So, for any positive real number $\epsilon$, there exists a positive real number $\delta_1$ such that $$0 < h < \delta_1 \implies |\int_{a + h}^{c} f - B| < \frac{\epsilon}{2}.$$
$$\lim_{h' \to 0+} \int_{c}^{b-h'} f = C.$$
So, for any positive real number $\epsilon$, there exists a positive real number $\delta_2$ such that $$0 < h' < \delta_2 \implies |\int_{c}^{b-h'} f - C| < \frac{\epsilon}{2}.$$
So, $$0 < h < \delta_1, 0 < h' < \delta_2 \implies |\int_{a+h}^{b-h'} f - (B + C)| \leq |\int_{a + h}^{c} f - B| + |\int_{c}^{b-h'} f - C| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$