Prove that the improper integral is independent of the intermediate value

Question

Suppose the function $ f:[a,b] \rightarrow \mathbb{R} $ only has singularities at $ a $ and $ b $, and the integral $$ \int_{a+\epsilon}^{b-\epsilon} f $$ exists for all $ \epsilon > 0 $. With the improper integral defined as $$ \int_a^b f = \int_a^c f + \int_c^b = \lim_{\epsilon\rightarrow 0^+} \int_{a+\epsilon}^c f + \lim_{\epsilon\rightarrow 0^+} \int_c^{b-\epsilon} f $$ for some $ c \in (a,b) $, prove that the improper integral does not change if $ c $ is replaced by some other choice $ d \in (a,b) $. That is, it is either undefined for both $ c $ and $ d $, or defined in both cases with the same integral value?

Answer 1

$$ \lim_{\epsilon\rightarrow 0^+} \int_{a+\epsilon}^c f= \lim_{\epsilon\rightarrow 0^+} \int_{a+\epsilon}^d f + \int_{d}^cf$$

$$ \lim_{\epsilon\rightarrow 0^+} \int_c^{b-\epsilon}f= \int_{c}^df +\lim_{\epsilon\rightarrow 0^+} \int_d^{b-\epsilon} f$$

Upon adding both sides of above you get the result.