[Edited]
For a real-valued continuous function $f$ defined on a Lebesgue measurable dense subset of $[0,2]$, consider an integral $$ \int_{[0,1]}\frac{f(s)}{\sqrt{1-s}}ds. $$
My question is whether this integral can be determined or not.
Here the integral is handled in the sense of Lebesgue integral. For the reason, if the kernel, namely the denominator of integrand does not appear, then it is clear that this is determined. However, this has a singularity, i.e, improper integral and so I'm confusing. For example, I'm wondering how I should consider them when $f$ is not defined at $s=1$.
If some conditions are needed to determine this integral, I'm glad if you show for me.
In addition, since I don't understand a difference of integrals for between discontinuous functions and functions defined on dense subset, I'm great happy if you teach me if possible.
Thank you in advance
[Motivation]
*This involves my future work.
For a upper semicontinuous function $f:[0,2]\to\mathbb{R}$, I want to interpret the following integral in a weak sense in the below way $$ \int_{[0,1]}\frac{f'(s)}{\sqrt{1-s}}ds. $$
- Consider the subdifferential $D^{+}f(t)$ for $t\in[0,2]$.
- Define a function $g:[0,2]\to\mathbb{R}$ by $g(t)=a\in D^{+}f(t)$.
- Interpret the above integral as $$ \int_{[0,1]}\frac{g(s)}{\sqrt{1-s}}ds $$
Now, there are some problems:
- $D^{+}f$ is dense in $[0,2]$ but is possibly empty and multivalued.
- Thus $g$ may not be defined at some points.
Please temporally assume conditions written in the beginning.