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I'm struggling in getting the intuition of two statements about measure theory:

Given a measure space $(X,F,\mu)$, $f \in M^+ $, where $M^+ = M^+(F) $ is the set of non negative F-measurable functions $f: X \rightarrow [0,\infty]. $

  1. $\mu(E) = 0 \Rightarrow \mu_f(E) = \int_E f d\mu = 0$

  2. $\int_X f d\mu = 0 \Rightarrow \mu (\{x \in X: f(x) \neq 0\}) = 0$

About the proof, we did as follow:

(First statement) Take $\{\varphi_n\} \in B_0^+$ (where $B_0^+$ is the set of all simple nonnegative F measurable functions on X) s.t. $\varphi_n \uparrow f$ (hence $f$ is measurable) so that $\varphi_n 1_E \uparrow f 1_E$. Let $\varphi_n = \sum_{i=1}^{K_n}\alpha_{i,n}1_{A_{i,n}}$. Then

$\mu_f(E) = \int_X f1_E d\mu$ = $lim_n \int_X \varphi_n 1_E d\mu $ (by monotone convergence)

$= lim_n\int_x \varphi_n1_Ed\mu =lim_n\int_x \sum_{i=1}^{K_n}\alpha_{i,n}1_{A_{i,n}\cap E}d\mu $ (by definition of $\varphi_n$ and by property of indicator function)
$=lim_n \sum_{i=1}^{K_n}\alpha_{i,n}1_{A_{i,n}\cap E} = 0$ since $0 \leq \mu(A_{i,n} \cap E) \leq \mu(E) = 0 $ (by assumption and by the fact that $A_{i,n} \subset E$).

(Second statement)

Let $E = \{ x \in X : f(x) \ne 0\} = \{x \in X: f(x) >0 \}$
For any $n \geq 1$, set $E_n = \{x \in X: f(x) \geq 1/n\}$, then $E_n \uparrow E$ and :
$0 = \int_X f d\mu \geq \int_Xf1_{E_n} d\mu \geq \int_{E_n}1/n d \mu = \frac{1}{n}\mu(E_n)\geq0 $
hence $\mu(E_n) = 0$ for all $n$. Therefore $0 \leq \mu(E) = \mu(\cup_{n=1}^{\infty} E_n) \leq \sum_{n = 1}^\infty \mu(E_n) = 0$

Could you please help me in getting some intuition?

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  • $\begingroup$ The second one is wrong without some additional assumptions such as $f$ being nonempty. $\endgroup$ Feb 23 at 9:18
  • $\begingroup$ That should be nonnegative. $\endgroup$ Feb 23 at 9:41
  • $\begingroup$ Yeah sorry, i forgot about the assumptions. Editing right now $\endgroup$ Feb 23 at 10:43
  • $\begingroup$ Do you understand the proofs of the results? There are different roads to these fundamental results and these roads might correspond to different intuitions. $\endgroup$ Feb 23 at 11:00
  • $\begingroup$ Yeah i get all the steps to get (i think) the two proofs but i cannot see the intuition behind the statements $\endgroup$ Feb 23 at 12:24

1 Answer 1

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For the first statement, note that $\int_E f~\mathrm d\mu$ is by definition the same thing as $\int 1_E f~\mathrm d\mu$, and the integral of a function is defined as the limit of the integrals of approximating simple functions. If $\langle \phi_n\rangle$ is a sequence of simple functions increasing pointwise to $f$, then $\langle\phi_n 1_E\rangle$ is a sequence of simple functions increasing pointwise to $1_E f$. Now the important point is that a simple function that is zero outside the measure zero set $E$ must have integral zero. It is a finite sum of "rectangles" whose bases are measure zero subsets of $E$. By the definition of the integral, the limit $\int 1_E f~\mathrm d\mu$ must be zero, too.

For the second statement, one really proves (a strengthening of) the contrapositive: If $f$ is larger than zero on a set of positive measure, then $f$ must have an integral larger than zero. There are really two steps for that. First, one proves that we must have that for some $n$, $f$ is at least $1/n$ on a set of positive measure $E_n$. Then, one proves that $\int f~\mathrm d\mu\geq 1/n~\mu_n(E_n)>0.$ This second step is easier and uses just the monotonicity of the integral for nonnegative functions, which follows essentially by definition. So $\int f~\mathrm d\mu\geq\int_{E_n} f~\mathrm d\mu$ because $f$ is noonegative. But for all $x\in E_n$, we have $f(x)\geq 1/n$. So $\int_{E_n} f~\mathrm d\mu \geq \int_{E_n} 1/n~\mathrm d\mu=1/n~\mu_n(E_n)$. So, let's do the first step that actually uses $\sigma$-additivity. Since for every strictly positive number $r$, there is some natural number $n$ satisfying, $1/n<r$m we must have $E=\bigcup_{n=1}^\infty E_n$. Moreover, if $m<n$, then $r<1/m$ implies $r<1/n$ and, therefore, $E_m\subseteq E_n$. So the sequence $\langle E_n\rangle$ is increasing (under set inclusion) with union $E$. The $\sigma$-additivity of $\mu$ implies then that $\mu(E)=\lim_n \mu(E_n)$. In particular, if $\mu(E)>0$, then we must have $\mu(E_n)>0$. And that is what we tried to show.

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  • $\begingroup$ Thanks a lot for the clarification!! $\endgroup$ Feb 23 at 17:45

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