# Intuition of two Measure theory statements

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 $$0 \leq \mu(E) = \mu(\cup_{n=1}^{\infty} E_n) \leq \sum_{n = 1}^\infty \mu(E_n) = 0$$

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

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.