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A well-known fact is that if a space $X$ is reflexive, then Dunford and Pettis integrals coincide. But I'm inerested in the relation between Bochner and Pettis integral. It seems obvious that if a function is Bochner integrable, then it is Pettis integrable as well. What are the necessary conditions so that they coincide?

I thought that $X$ being separable is enough. Then the function we integrate becomes strongly measurable. Is it enough, or maybe separability is too much?

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    $\begingroup$ If a function is Bochner integrable then its Bochner integral is equal to its Pettis integral. Is this what you are asking? $\endgroup$ Commented Apr 12, 2019 at 10:07
  • $\begingroup$ Yes, and also, a function is Pettis integrable plus propably something else (what is that?) is Bochner integrable. $\endgroup$
    – zorro47
    Commented Apr 12, 2019 at 10:09
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    $\begingroup$ Bochner integrabilty is equivalent to strong measurability and finiteness of $\int \|f\|$. There is no other simpler condition under Pettis integrability. $\endgroup$ Commented Apr 12, 2019 at 10:14
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    $\begingroup$ So, if a function is Pettis integrable, then it is weakly measurable. And if a space is for instance separable, then $f$ becomes strongly measurable. If in addition, $\int\|f\|<\infty$ then it becomes Bochner integrable. $\endgroup$
    – zorro47
    Commented Apr 12, 2019 at 10:17
  • $\begingroup$ Exactly. That is the best you can say. $\endgroup$ Commented Apr 12, 2019 at 10:18

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