A function $f\in X^{[a, b]}$ is said to be Riemann-Pettis integrable if $x^* f$ is Riemann integrable for each $x^*\in X^*$ and if for each interval $I\subset [a, b]$, there is a vector in $x_I\in X$ such that $x^*(x_I)=\int_I x^* f(t) dt$ for all $x^*\in X^*$.
Is range of the induced vector measure of a Riemann-Pettis integrable function relatively compact in X?
Dear Pratikshan Mondal
It seems to me that the answer is positive. Indeed, every Riemann integrable function is bounded, so for each $x^*\in X^*$ the set $x^* f([a, b])$ is bounded. Then, by the Uniform Boundedness principle, $f([a, b])$ is bounded in norm. In other words, there is a constant $M > 0$ such that $||f(t)|| < M$ for all $t$.
Consider the function $F(t) = \int_a^t f(s)ds$. This function satisfies the Lipschitz condition with constant $M$, so it is continuous. The image of a compact set under a continuous map is compact, so $F([a, b])$ is compact. Finally, the vector measure in question is defined on the collection of all intervals of the form $[t_1, t_2]$, its values are $F(t_2) - F(t_1)$, so the range lies in the Minkowski sum $F([a, b]) + (-F([a, b]))$, which is compact.
Best, Vladimir
Dear Pratikshan Mondal and José Rodríguez ,
It is nice to see that real fans of mathematics continue doing math. even on December 31 :)
Remark, that my proof given above does not use any difficult argument and can be understood by anybody who knows basic Functional Analysis.
Best wishes in the coming New Year, Vladimir
Dear Vladimir Kadets and José Rodríguez,
Thank you both for your response. Yes, it is really nice to see discussing on a question on 31 st December. Happy New year 2020 to both of you.
@Rodriguez Please tell me a reference to find Stegall result. Thank you again.
@ Vladimir Kadets , Sir firstly a very happy new year.
Sir, in your answer you have used only the boundedness of the Riemann-Pettis integral and continuity of the indefinite integral. Therefore, is it true that a bounded Pettis integrable function has the same property? i.e.,
the range of the induced vector measure of a bounded Pettis integrable function relatively compact in X?
@Rodriguez Sir , a very happy new year.
The result you have used is for a perfect measure. Is it true that Lebesgue measure on the set of all Lebesgue measurable subsets of [a, b] is perfect?
Dear Pratikshan Mondal
Answering your question "is the range of the induced vector measure of a bounded Pettis integrable function relatively compact in X", I would remark that the answer is evidently positive if by "range" you mean the set of values on all segments [t_1, t_2] (like for the Riemann integral). If the "range" is the set of values on all measurable subsets, the answer is not so clear for me, and one needs more effort.
Best, Vladimir
Dear Vladimir Kadets, the range means the set of all values on all measurable subsets. I think the answer is negative in this case.
What do you think?
Dear Pratikshan Mondal ,
I don't know the answer. In separable case a bounded Pettis integrable function is Bochner integrable, so in this case we have precompactness of the range. In the standard example that I know that do not reduce to separable case the range is also precompact. But I don't know if this is true in general. Maybe José Rodríguez knows? He has much better education in the subject than I have.
Best, Vladimir
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