Integral of a function defined on dense subset

[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.

[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 $$\int_{[0,1]}\frac{f'(s)}{\sqrt{1-s}}ds.$$

1. Consider the subdifferential $D^{+}f(t)$ for $t\in[0,2]$.
2. Define a function $g:[0,2]\to\mathbb{R}$ by $g(t)=a\in D^{+}f(t)$.
3. Interpret the above integral as $$\int_{[0,1]}\frac{g(s)}{\sqrt{1-s}}ds$$

Now, there are some problems:

1. $D^{+}f$ is dense in $[0,2]$ but is possibly empty and multivalued.
2. Thus $g$ may not be defined at some points.

Please temporally assume conditions written in the beginning.

• There is a problem: the dense subset, which is the domain of $f$, may not be Lebesgue measurable. In such a case, the integral is not even defined. – Crostul Mar 31 '16 at 21:31
• even if $f$ is defined on the whole interval and there is no denominator, the integral need not exist. Do you assume that $f$ is measurable, what else do you assume? – Mirko Mar 31 '16 at 21:33
• Why do you want a dense set? – mlainz Mar 31 '16 at 21:37
• Maybe by "functions defined on a dense subset" you are talknig about "functions defined almost everywhere"? These two concepts are not equivalent. – Crostul Mar 31 '16 at 21:38
• @Crostul Yes, I know that. I forget to write it and please think that it is assumed. – user Mar 31 '16 at 21:43