Can someone give an example why a non-complete measure is a problem? Or what are some theorems that require a measure to be complete?
38465
While reading the motivation of complete measure space , what I concluded was,  completeness is not really necessary when we define on one measure space  and it is necessary when we want to measure on product of measure spaces (is it true ?).  If $\lambda$ is measure on $X$ and $Y$ then is it true that $\lambda^2$ is measure of $A$x$B$ and how ?  I am not able to understand that $\lambda^2(A\times B)=\lambda(A)\times\lambda(B)$ ?  Essentially what is the flaw in the measure without being complete ?  Waiting for response. Thanks!
26172
As a problem during introduction to measure theory, I am asked to prove the following.  Let $A \subset [0,1]$ such that $A$ is measurable with $\mu(A) = 0$ ($\mu$ being the Lebesgue measure). Prove that $A^c =  [0,1] \setminus A$ is dense in $[0,1]$.  This makes intuitive sense to me $-$ since $\mu(A) = 0$, using that $\mu(X) = \mu(X \cap A) + \mu(X \cap A^c)$ for every $X \subseteq \mathbb{R}$ we get $\mu([0,1]) = \mu(A) + \mu(A^c) = \{\mu(A)=0\} = \mu(A^c)$. Thus, $A^c$ has the same measure as $[0,1]$. I am aware that this is not the definition of $A^c$ being dense in $[0,1]$ though, and I do not know how to formalize this proof.  Any advice would be appreciated.
61711
This is an interesting question that has stumped the entirety of my measure theory class, including the professor:  Prove or disprove:  Let $(X,\mathcal A)$, $(Y,\mathcal B)$ be measure spaces.  Let $f$ be an $\mathcal A$ $\times$ $\mathcal B$ measurable nonnegative function, and let $g(x)$ $=$ $sup_{y \in Y}f(x,y)$, with $g(x)&lt;\infty$ for all $x$. Is $g(x)$ necessarily an $\mathcal A$-measurable function?  We all feel the answer is no, given that slices are measurable, and sups of measurable functions are only guaranteed to be measurable over a countable index. We think the correct answer is to start with a nonmeasurable set $S$ in $X$, and to try to build a set $T$ in $Y$ that makes $S \times T$ measurable in $\mathcal A \times \mathcal B$, but we suspect this is quite difficult with no further guidance.   Any ideas?
37502
Is it possible to construct a non-measurable set in $[0,1]$ of a given outer measure $x \in [0,1]$? This will probably require the axiom of choice. Does anyone have a suggestion?  Edit: I forgot to mention that the set has to be in $[0,1]$ which seemed "clear" in my head, but which obviously isn't.  Edit 2: Now it is clear that I just need a non-measurable set with outer measure 1, from that I can make any other one. I have once seen somewhere that they do this with transfinite recursion (if I recall correctly), but I don't remember where and how because I didn't know that transfinite recursion was back then.
111184
$m$ is the Lebesgue measure. I was thinking that: $\Bbb Q$ is dense and $m(\Bbb Q)=0&lt;1$ but it fails to be open, but maybe I could construct an open set from this fact, or also using that $(0,1)$ is a dense and open subset but $m(0,1)=1$.   Any hints or ideas will be very appreciated. Thanks.
25689
Is the motivation behind the introduction of measure theory the Lebesgue integral? In order to evaluate such an integral we need the length of each of the horizontal strip of width $h$. I have a question here. Why do we need such a general concept of measure? Length is a kind of measure, but why did Lebesgue think of so general a concept of "measure" when he just needed the length to evaluate his integrals? Is there any other motivation behind this ?
75089
If $\mu$ is a Radon measure on $\mathbb{R}^n$ and $B_r$ is a closed ball of radius $r$. Why is $\mu(\partial B_r) = 0$? Or how can I prove that there is at least one $r_0 &gt; 0$ such that $\mu(\partial B_{r_0}) = 0$?
123902
I have seen that completeness is not a topological property like compactness or connectedness. I have seen some examples also showing that there are two equivalent metrics one of which is complete and the other one is incomplete. I want to know some general result. Consider any metric space $(X,d)$. Let $d$ be complete. Does there exist an equivalent metric $d'$ in $X$ which is incomplete? When $d$ is compact, such $d'$ does not exist (Since compactness implies completeness). So answer me when $d$ is noncompact.
83306
Hello all mathematicians!! Again, I am struggling with solving the exercises in Lebesgue Integral for preparing the quiz. At this moment, I and my friend are handling this problem, but both of us agreed this problem is a bit tricky. The problem is following.   Let ($X,\mathcal{A},\mu$) be a measure space and suppose $\mu$ is $\sigma$-finite. Suppose $f$ is integrable. Prove that given $\epsilon$ there exist $\delta$ such that   $$ \int_A |f(x)|\mu(dx) \;\;&lt;\;\;\epsilon $$  whenever $\mu(A) &lt; \delta$.  Could anybody give some good idea for us? Think you very much for your suggestion in advance.
77932
I am trying to solve following exercise from Folland,     If $\mu$ is a semifinite measure and $\mu(E) = \infty$, for any $C &gt; 0$, $\exists$ $F \subset E$ with $C &lt; \mu(F) &lt; \infty$.   It seems to follow from definition of semifinite measures, which you can find , but I couldn't prove it.
87209
this is a follow up to the problem described .  I have the same problem and the proposed solution does not completely solve the problem. I mean, it does solve the problem but also creates another one. The new problem is that the theorems, lemmas and the rest are placed in such as way that they are not cut (of course), but this leaves then big empty spaces in many pages. So the problem is that, the problem described  does not seem to be fully resolved using the nocut option.   Do you any of you have a good idea on how to solve this?  Thanks!
45022
Question: Let $E_1,E_2,...,E_n$ be measurable subsets of $[0,1]$ with $$ \sum_{k=1}^n m(E_k)  &gt; n-1.$$ Prove that $\bigcap_{k=1}^n E_k$ has positive measure.  This is one of the questions in graduate analysis past year paper. I think we need to assume that the intersection $\bigcap_{k=1}^n E_k$ is nonempty. Otherwise, the question is false. Anyway, let's assume that $\bigcap_{k=1}^n E_k \neq \emptyset.$ If $n=2,$ by inclusion-exclusion principle and assumption, $$m(E_1\cup E_2) + m(E_1 \cap E_2) = m(E_1) + m(E_2) &gt; 2.$$ Since $E_1,E_2 \subseteq [0,1],$ by monotonicity of Lebesgue measure, $$m(E_1) \leq 1, m(E_2) \leq 1.$$ By finite subadditivity of Lebesgue measure, $$m(E_1 \cup E_2) \leq m(E_1) + m(E_2) \leq 2.$$ Hence, $$m(E_1 \cap E_2) &gt; 2 - m(E_1 \cup E_2) \geq 0$$ Therefore, the intersection $\bigcap_{k=1}^n E_k$ has positive measure for $n=2.$ I try prove the general $n$ using induction, but it seems a bit long. Does there exist an efficient method to solve the question? EDIT: Actually I am looking for a direct proof instead of indirect proof or inductive proof. If a direct proof is not possible, then I will accept Marios's answer.
36431
Ran across this problem and need some help.   Let $E$ be a measurable subset of the real numbers and suppose that for any open interval $I$ one has $m(E\cap I)\le \frac{1}{2}m(I)$, where $m$ is the Lebesgue measure. Prove that $m(E)=0$.
40873
I know it is true that every countable set has measure zero, but is the converse true. Is it true that every set of measure zero is countable?
71579
I would like to teach myself measure theory. Unfortunately most of the books that I've come across are very difficult and are quick to get into Lemmas and proofs. Can someone please recommend a layman's guide to measure theory? Something that reads a bit like this , starts out very gently and places much emphasis on the intuition behind the subject and the many lemmas.
50405
I'm trying to prove the following statements in Folland's book.  Let $(X,\mathcal{M},\mu)$ be a measure space. If $f_n\to f$ in measure and $g_n\to g$ in measure, then $f_n+g_n\to f+g$ in measure and $f_ng_n\to fg$ in measure (in this last case, assume $\mu(X)&lt;\infty$).  The first one is already done, but the second one is giving me trouble.   My attempt: We can take subsequences $f_{n_k}$ and $g_{n_k}$ such that $f_{n_k}\to f$ a.e. and $g_{n_k}\to g$ a.e. In this case, we know that $f_{n_k}g_{n_k}\to fg$ a.e. Note that $\mu(X)&lt;\infty$, so we can use Egoroff's theorem to conclude that $f_{n_k}g_{n_k}\to fg$ almost uniformly. Finally, almost uniform convergence implies convergence in measure, therefore $f_{n_k}g_{n_k}\to fg$ in measure.   The problem is that I just have convergence in measure for some subsequence of $f_ng_n$ and I need convergence in measure for the whole sequence. Also I don't know if using Egoroff's theorem is unnecessary, if this is the case, I would appreciate to see some more elementary prove.  Detail: We say that $f_n\to f$ almost uniformly if for every $\varepsilon &gt; 0$ there is some $E\subset X$ measurable, such that $\mu(E) &lt;\varepsilon$ and $f_n\to f$ uniformly on $E^c$.  Thank you!
93689
This question was asked , but the question wasn't asking for an answer, just a clarification on the proof by contradiction they already had, so I'm reposting it to ask for help arriving at the answer.  Because the asker above said they proved it by contradiction, that is what I'm attempting. But I'm having issues wrapping my brain around the concept of set measurability. So let's give it a good college try with what I can understand.  For clarity, $\mu$ is an outer measure.  A is nonmeasurable, so $\exists B \subseteq X$ such that $\mu (B) \ne \mu (B \bigcap A) + (B \bigcap A^c)$.  E is measurable, so $\mu (B) = \mu (B \bigcap E) + (B \bigcap E^c)$.  This essentially means $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$ (Since monotonicity is alredy implied by the definition of outer measure).  So we can attempt a proof (or the start of one) by contradiction.  Assume $\mu (E \setminus A)=0$. So $E \setminus A$ is a null set.  So $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$. I assume I'm supposed to show a contradiction using that A is nonmeasurable. I'm unsure how.  Another way I think could work would be to use: $\mu (E \setminus A)= \mu (E \bigcap A^c) = 0$ and arrive at a contradiction given that A is nonmeasurable, but again, I'm unsure how.  I just need some direction from here. Thank you in advance for any advice.