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I was studying the answers to the following question: About two functions whose Lebesgue integral on all sets of a $\sigma-$algebra are equal. Now I am wondering how to interpret the sets $\{f>g\}$, $\{f=g\}$,... and so on.

In the case where $f$ and $g$ are continuous (and thus Borel-measurable), the case is clear and we can write for example $\{f>g\}=\{x\in X\colon f(x)>g(x)\}$.

But what about the case where $f,g\in L^p(X)$ are measurable? I learned that elements of $L^p$-spaces are equivalence classes of measurable and integrable functions. So how can we define such a set, when the values of some function $f$ in the equivalence class $[f]$ are arbitrary in certain (or even every) points? If I choose representatives $f$ and $g$ and try to determine the set $\{f>g\}$, can't it become a completely different set, if I choose new representatives?

I always become completely confused every time when it comes to $L^p$-functions and pointwise arguments. Can you give me some simple or robust intuition?

bongobums
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  • I would interpret ${f>g}$ as $f-g>0$ a.e. (for an arbitrary representative) . The latter should make sense because representatives only differ on null sets. – Hyperbolic PDE friend Jul 16 '20 at 09:35

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The meaning of $[f]>[g]$ is that $f,g$ are almost everywhere real valued, and almost everywhere we have $f(x)>g(x)$. This doesn't depend on the choice of the representatives, because if you change the representatives you will change the functions only on a set of measure zero.

Mark
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  • But the set ${f>g}$ can contain any point of $X$? What I am just imagining is that, to determine the set, I take any two representatives of the equivalence class and compare them. And the set I get then is exactly the same set (except for a set with measure zero) that I would get for two other representatives? – bongobums Jul 16 '20 at 10:20
  • Can you write down the set I am looking for explicitly please? In the form I did for the continuous functions? – bongobums Jul 16 '20 at 10:23
  • Almost all points have to belong to the set ${x\in X: f(x), g(x)\in\mathbb{R}, \ \ f(x)>g(x)}$. (I'm assuming $f,g$ are complex valued functions. If you assume they are real valued you may not write $f(x),g(x)\in\mathbb{R}$) – Mark Jul 16 '20 at 10:42
  • Okay, just to be clear. You take two representatives $f$ and $g$ and define the above set. Then the set ${f>g}$ (which is related to the equivalence classes $[f]$ and $[g]$) is almost the same as the set above, i.e. they differ only on a set of measure zero, which we do not care about since we are integrating. Is that right? – bongobums Jul 16 '20 at 10:51
  • Yes, if you change the representatives this is almost the same set, the difference will be only on a set of measure zero. And a change in a set of measure zero doesn't change the integral. – Mark Jul 16 '20 at 11:18