<p>For the first statement, note that <span class="math-container">$\int_E f~\mathrm d\mu$</span> is by definition the same thing as <span class="math-container">$\int 1_E f~\mathrm d\mu$</span>, and the integral of a function is defined as the limit of the integrals of approximating simple functions. If <span class="math-container">$\langle \phi_n\rangle$</span> is a sequence of simple functions increasing pointwise to <span class="math-container">$f$</span>, then <span class="math-container">$\langle\phi_n 1_E\rangle$</span> is a sequence of simple functions increasing pointwise to <span class="math-container">$1_E f$</span>. Now the important point is that a simple function that is zero outside the measure zero set <span class="math-container">$E$</span> must have integral zero. It is a finite sum of &quot;rectangles&quot; whose bases are measure zero subsets of <span class="math-container">$E$</span>. By the definition of the integral, the limit <span class="math-container">$\int 1_E f~\mathrm d\mu$</span> must be zero, too.</p>
<p>For the second statement, one really proves (a strengthening of) the contrapositive: If <span class="math-container">$f$</span> is larger than zero on a set of positive measure, then <span class="math-container">$f$</span> must have an integral larger than zero. There are really two steps for that. First, one proves that we must have that for some <span class="math-container">$n$</span>, <span class="math-container">$f$</span> is at least <span class="math-container">$1/n$</span> on a set of positive measure <span class="math-container">$E_n$</span>. Then, one proves that <span class="math-container">$\int f~\mathrm d\mu\geq 1/n~\mu_n(E_n)&gt;0.$</span> This second step is easier and uses just the monotonicity of the integral for nonnegative functions, which follows essentially by definition. So
<span class="math-container">$\int f~\mathrm d\mu\geq\int_{E_n} f~\mathrm d\mu$</span> because <span class="math-container">$f$</span> is noonegative. But for all <span class="math-container">$x\in E_n$</span>, we have <span class="math-container">$f(x)\geq 1/n$</span>. So <span class="math-container">$\int_{E_n} f~\mathrm d\mu \geq \int_{E_n} 1/n~\mathrm d\mu=1/n~\mu_n(E_n)$</span>. So, let's do the first step that actually uses <span class="math-container">$\sigma$</span>-additivity. Since for every strictly positive number <span class="math-container">$r$</span>, there is some natural number <span class="math-container">$n$</span> satisfying, <span class="math-container">$1/n&lt;r$</span>m we must have <span class="math-container">$E=\bigcup_{n=1}^\infty E_n$</span>. Moreover, if <span class="math-container">$m&lt;n$</span>, then <span class="math-container">$r&lt;1/m$</span> implies <span class="math-container">$r&lt;1/n$</span>  and, therefore, <span class="math-container">$E_m\subseteq E_n$</span>. So the sequence <span class="math-container">$\langle E_n\rangle$</span> is increasing (under set inclusion) with union <span class="math-container">$E$</span>. The <span class="math-container">$\sigma$</span>-additivity of <span class="math-container">$\mu$</span> implies then that <span class="math-container">$\mu(E)=\lim_n \mu(E_n)$</span>. In particular, if <span class="math-container">$\mu(E)&gt;0$</span>, then we must have <span class="math-container">$\mu(E_n)&gt;0$</span>. And that is what we tried to show.</p>

CC BY-SA 4.0
2024-02-23T17:45:06.983Z
Thanks a lot for the  clarification!!
Calogero Sortino
CC BY-SA 4.0
2024-02-23T14:28:32.590Z
2024-02-23T14:28:32.590Z
Michael Greinecker
<p>I'm struggling in getting the intuition of two statements about measure theory:</p>
<p>Given a measure space <span class="math-container">$(X,F,\mu)$</span>, <span class="math-container">$f \in M^+ $</span>, where <span class="math-container">$M^+ = M^+(F) $</span> is the set of non negative F-measurable functions <span class="math-container">$f: X \rightarrow [0,\infty]. $</span></p>
<ol>
<li><p><span class="math-container">$\mu(E) = 0 \Rightarrow \mu_f(E) = \int_E f d\mu = 0$</span></p>
</li>
<li><p><span class="math-container">$\int_X f d\mu = 0 \Rightarrow \mu (\{x \in X: f(x) \neq 0\}) = 0$</span></p>
</li>
</ol>
<p>About the proof, we did as follow:</p>
<p><strong>(First statement)</strong>
Take <span class="math-container">$\{\varphi_n\} \in B_0^+$</span> (where <span class="math-container">$B_0^+$</span> is the set of all simple nonnegative F measurable functions on X) s.t. <span class="math-container">$\varphi_n \uparrow
 f$</span> (hence <span class="math-container">$f$</span> is measurable) so that <span class="math-container">$\varphi_n 1_E \uparrow f 1_E$</span>. Let <span class="math-container">$\varphi_n = \sum_{i=1}^{K_n}\alpha_{i,n}1_{A_{i,n}}$</span>. Then</p>
<p><span class="math-container">$\mu_f(E) = \int_X f1_E d\mu$</span>  = <span class="math-container">$lim_n \int_X \varphi_n 1_E d\mu $</span> (by monotone convergence)<br/></p>
<p><span class="math-container">$= 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 $</span> (by definition of <span class="math-container">$\varphi_n$</span> and by property of indicator function) <br/>
<span class="math-container">$=lim_n \sum_{i=1}^{K_n}\alpha_{i,n}1_{A_{i,n}\cap E} = 0$</span> since <span class="math-container">$0 \leq \mu(A_{i,n} \cap E) \leq \mu(E) = 0 $</span> (by assumption and by the fact that <span class="math-container">$A_{i,n} \subset E$</span>).</p>
<p><strong>(Second statement)</strong></p>
<p>Let <span class="math-container">$E = \{ x \in X : f(x)  \ne 0\} = \{x \in X: f(x) &gt;0 \}$</span> <br/>
For any <span class="math-container">$n \geq 1$</span>, set <span class="math-container">$E_n = \{x \in X: f(x) \geq 1/n\}$</span>, then <span class="math-container">$E_n \uparrow E$</span> and : <br/>
<span class="math-container">$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 $</span> <br/>
hence <span class="math-container">$\mu(E_n) = 0$</span> for all <span class="math-container">$n$</span>. Therefore <span class="math-container">$0 \leq \mu(E) = \mu(\cup_{n=1}^{\infty} E_n) \leq \sum_{n = 1}^\infty \mu(E_n) = 0$</span></p>
<p>Could you please help me in getting some intuition?</p>

CC BY-SA 4.0
2024-02-23T09:18:43.153Z
The second one is wrong without some additional assumptions such as $f$ being nonempty.
Michael Greinecker
CC BY-SA 4.0
2024-02-23T09:41:45.517Z
That should be nonnegative.
Michael Greinecker
CC BY-SA 4.0
2024-02-23T10:43:11.673Z
Yeah sorry, i forgot about the assumptions. Editing right now
Calogero Sortino
CC BY-SA 4.0
2024-02-23T11:00:43.887Z
Do you understand the proofs of the results? There are different roads to these fundamental results and these roads might correspond to different intuitions.
Michael Greinecker
CC BY-SA 4.0
2024-02-23T12:24:13.083Z
Yeah i get all the steps to get (i think) the two proofs but i cannot see the intuition behind the statements
Calogero Sortino
CC BY-SA 4.0
2024-02-23T13:16:46.697Z
What intuition could there be with sigma subadditivity, where summing up zero measures you sometimes get a positive measure... do you have intuitive stories where you deal with countably infinite sequences?
Giskard
<mathematical-economics><proof><proof-reference>
Intuition of two Measure theory statements
Calogero Sortino
# Intuition of two Measure theory statements

(Asked by: username_0 on 2024-02-23)

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?

**Tags:** mathematical-economics, proof, proof-reference
> **Comments:**
> **username_1**: The second one is wrong without some additional assumptions such as $f$ being nonempty.  
> **username_1**: That should be nonnegative.  
> **username_0**: Yeah sorry, i forgot about the assumptions. Editing right now  
> **username_1**: Do you understand the proofs of the results? There are different roads to these fundamental results and these roads might correspond to different intuitions.  
> **username_0**: Yeah i get all the steps to get (i think) the two proofs but i cannot see the intuition behind the statements  
> **username_2**: What intuition could there be with sigma subadditivity, where summing up zero measures you sometimes get a positive measure... do you have intuitive stories where you deal with countably infinite sequences?  

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## 1 answers

### Answer by username_1 on 2024-02-23. Score: 3

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/nm we must have $E=\bigcup\_{n=1}^\infty E\_n$. Moreover, if $m, 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.

> **Comments:**
> **username_0**: Thanks a lot for the  clarification!!  

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