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I want to prove Egorov's theorem using this Lebesgue integral defined by the upper integral $$\int^*f:=\left\{\int h ; h \ge f \text{ and h upper-continuous }\right\}$$

$$\int_*f:=\left\{\int h ; h \le f \text{ and h lower-continuous }\right\}$$

So a Lebesgue integral of a function $ f : \mathbb{R}^n \rightarrow \mathbb{R}$ exists $\int f \Leftrightarrow \int^*f = \int_*f$.

I am also allowed to use the following theorems:

  • $L^p$ is a Banach space;
  • Dominated convergence theorem;
  • Monotone convergence theorem;
  • $C^{\infty}$ is dense in the Lebesgue-functions.

But the huge problem is: We don't know what Borel-sets are and we don't have anything like measures so far. Therefore, all standard proofs of this theorem are not applicable to this situation. Hence, I wanted to find out whether anybody here knows a way how to do it?

Maybe I should say more about how this integral is defined:

Every semincontinuous function is the limit of a monotone sequence of continuous functions with finite support $g_n$. The integral over these kind of functions is defined via n-times 1 dimensional integration over all variables and then $\int h:= \lim_{n \rightarrow \infty} \int g_n$. (But this is probably not that relevant to this proof).

If anything is unclear, please let me know

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  • 4
    $\begingroup$ How do you state Egorov's theorem without referring to a measure? $\endgroup$ – Justthisguy Jan 24 '14 at 18:58
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    $\begingroup$ The measure of a set is just defined as $\mu(A):= \int_{\mathbb{R}^n} 1_A(x)dx$ $\endgroup$ – user66906 Jan 25 '14 at 16:18
  • 4
    $\begingroup$ Still, could you write explicitely the statement that you want to prove? Usually, Egorov involves measurable functions... $\endgroup$ – Etienne Jan 31 '14 at 8:15
  • $\begingroup$ Yes, so it stays the same, cause a function in this case is called measurable iff the following integral exists( integral from above and below coincide) and we have that $|\int_{\mathbb{R}^n} f | < \infty$ $\endgroup$ – user66906 Feb 1 '14 at 10:49
  • $\begingroup$ The integral in your definition of measure $\mu$ is the Riemann-integral? If yes, then the sufficient and necessary condition of R-integrability gives the description of those sets $A$ for which you can define $\mu$. After that you should verify that $\mu$ is a measure at all. $\endgroup$ – vesszabo Feb 1 '14 at 14:18

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