This is probably a duplicate of some other question, but it's not immediately obvious which. The fat cantor set is constructed by removing smaller fractions of the center in each stage of the cantor set construction, and it is claimed that this set has positive measure. To be precise, let $C_0=[0,1]$ and $C_n$ be the result of removing the central $1/2^{n+2}$-th of each interval in $C_{n-1}$, and let $C=\bigcap_{n=0}^\infty C_n$. Then $\mu(C_n)=\frac12(1+2^{-n})$, and as a countable intersection of measurable sets $C$ is measurable. Since each $C_n$ is a countable (finite, even) interval cover of $C$, we have $\mu(C)\le\frac12$. But how does one prove $\mu(C)\ge\frac12$? The only proof I have seen that bounds a measure away from zero is the proof that $\mu([a,b])=b-a$, and it's not a simple proof. What argument gives this lower bound on $\mu(C)$? Obviously I can't use monotonicity since $C$ has no interior.
[Math] Fat cantor set has positive lebesgue measure
lebesgue-measuremeasure-theoryreal-analysis
Related Solutions
Let $\{I_{k,j}:j=1,\ldots,2^k\}$ be the set of open intervals we are throwing away on $k$-th step of our construction. Note that $m(I_{k,j})=5^{-k-1}$ for all $k\in\mathbb{N}$ and $j=1,\ldots,2^k$.
Let $$ C_0:=[0,1]\qquad\qquad C_{k+1}:=C_k\setminus\bigcup\limits_{j=1}^{2^k}I_{k,j} $$ Now we proceed to the proof of inequality. Define $f_n=\chi_{C_n}$, then for $m\geq n$ we have $$ \int_0^1|f_m(x)-f_n(x)|dm= \int_0^1|\chi_{C_m}(x)-\chi_{C_n}(x)|dm= \int_0^1\chi_{C_n\setminus C_m}(x)dm= m(C_n\setminus C_m)= m\left(\bigcup\limits_{k=n}^{m-1}\bigcup\limits_{j=1}^{2^k} I_{k,j}\right)= \sum\limits_{k=n}^{m-1}\sum\limits_{j=1}^{2^k}m(I_{k,j})= \sum\limits_{k=n}^{m-1}\sum\limits_{j=1}^{2^k}5^{-k-1}= \sum\limits_{k=n}^{m-1}2^k5^{-k-1} $$ Hence $$ \sup\limits_{m\geq n}\int_0^1|f_m(x)-f_n(x)|dm= \sup\limits_{m\geq n}\sum\limits_{k=n}^{m-1}2^k5^{-k-1}= \sum\limits_{k=n}^\infty 2^k5^{-k-1}=\frac{2^n 5^{-n-1}}{1-2\cdot 5^{-1}}=\frac{1}{3}\left(\frac{2}{5}\right)^n $$
We proceed to the second part. Define $$ C:=\bigcap\limits_{k=1}^\infty C_k $$ Its measure equals $$ m(C)= 1-m(C_0\setminus C)= 1-m\left(\bigcup\limits_{k=0}^\infty\bigcup\limits_{j=1}^{2^k}I_{k,j}\right)= 1-\sum\limits_{k=0}^\infty \sum\limits_{j=0}^{2^k} m(I_{k,j})=\\ 1-\sum\limits_{k=0}^\infty \sum\limits_{j=0}^{2^k} 5^{-k-1}= 1-\sum\limits_{k=0}^\infty 2^k 5^{-k-1}= 1-\frac{5^{-1}}{1-2/5}=\frac{2}{3} $$ Now note that $x\in C$ iff its representation in a fivefold number system have no digit $2$ at any place. Hence for a given $x\in C$ you can change some digit to $2$ to leave the set $C$. Moreover you can change digit with very big place number to get a number outside $C$ but very close to $x$. Thus for each $x\in C$ and each $\varepsilon>0$ there exist $\hat{x}\notin C$. This is equivalent to discontinuity of $f$ in point $x\in C$. So $f$ is discontinuous at any point of $C$.
Since $m(C)>0$ by Lebesgue critirion $f\notin\mathcal{R}([0,1])$, because set of discontinuities of $f$ have positive measure. Let $\hat{f}$ be a function equivalent to $f$. Assume $\hat{f}$ is Riemann interable, then it is bounded. Hence the function $\delta=\hat{f}-f$ is bounded. Since $\hat{f}$ is equivalent to $f$, then $\delta$ is non zero only on the set of zero measure. By Lebesgue criterion $\delta\in\mathcal{R}([0,1])$. Then $f=\hat{f}-\delta\in$$\mathcal{R}([0,1])$ as difference of two Riemann integrable functions. Contradiction, so $\hat{f}\notin\mathcal{R}([0,1])$
Hint: Think about Bass's Proposition 4.14 (1): given any $\delta > 0$, there is an open set $G$ with $A \subset G$ and $m(G - A) < \delta$. Then recall (or prove) that every open set is a countable union of disjoint open intervals.
Here are some more details. An argument like yours has difficulties with the possibility $m(A) = \infty$, but we can reduce to the case $m(A) < \infty$ by intersecting $A$ with large bounded sets, as below.
Suppose $\epsilon \in (0,1)$ is such that $m(A \cap I) \le (1-\epsilon)m(I)$ for each interval $I$. Let $k$ be a positive integer and let $A_k = A \cap [-k,k]$. Note that $m(A_k) \le 2k < \infty$, and that for each interval $I$, we have $$m(A_k \cap I) \le m(A \cap I) \le (1-\epsilon) m(I). \tag{1}$$
Let $\alpha > 0$ be arbitrary, and using Proposition 4.14, choose an open set $G$ with $A_k \subset G$ and
$$m(G - A_k) < \alpha \epsilon. \tag{2}$$
In particular, $m(G) = m(A_k) + m(G-A_k) < 2k + \alpha \epsilon < \infty$.
Now $G$ can be written as a disjoint union of open intervals: $G = \bigcup_{n=1}^\infty I_n$. Note that $m(I_n) \le m(G) < \infty$ for each $n$. Also, $G - A_k = \bigcup_{n=1}^\infty (I_n - A_k)$ which is also a disjoint union.
Now since $I_n - A_k = I_n - (I_n \cap A_k)$, we have
$$m(I_n - A_k) = m(I_n) - m(I_n \cap A_k) \ge m(I_n) - (1-\epsilon)m(I_n) = \epsilon m(I_n) \tag{3}$$ using (1). So by countable additivity,
$$m(G - A_k) = \sum_{n=1}^\infty m(I_n - A_k) \ge \sum_{n=1}^\infty \epsilon m(I_n) = \epsilon m(G). \tag{4}$$
Combining (1) and (3), we get $\epsilon m(G) < \alpha \epsilon$, so $m(G) < \alpha$. In particular, since $A_k \subset G$, we have $m(A_k) < \alpha$. But $\alpha > 0$ was arbitrary, so we must have $m(A_k) = 0$.
Moreover, $k$ was arbitrary, so $m(A \cap [-k,k]) = 0$ for every $k$. Since $A = \bigcup_{k = 1}^\infty (A \cap [-k,k])$, by countable additivity we conclude $m(A) = 0$.
Now let's drop the assumption $m(A) < \infty$. For any $n$ and any interval $I$, we have $$m(A \cap [-n,n] \cap I) \le m(A \cap I) \le (1-\epsilon)m(I).$$ Hence $A \cap [-n,n]$ satisfies the same condition and moreover $m(A \cap [-n,n]) \le 2n < \infty$. So by the previous case, $m(A \cap [-n,n]) = 0$. Now since $A = \bigcup_{n=1}^\infty (A \cap [-n,n])$, by countable additivity $m(A) = 0$.
For an explicit example with a fat Cantor set, let's consider the example given on Wikipedia, where at stage $n \ge 1$ we remove $2^{n-1}$ intervals, each of length $2^{-2n}$. The final set $C$ has measure $1 - \sum_{n=1}^\infty 2^{n-1} \cdot 2^{-2n} = 1 - \sum_{n=1}^\infty 2^{-n-1} = \frac{1}{2}$.
Let $I_k$ be the leftmost interval that remains after stage $k$. At stage $k$ the leftmost interval that was removed was centered at $2^{-k}$ and had length $2^{-2k}$, so the leftmost interval that remains is $$I_k = \left[0, 2^{-k} - \frac{1}{2} 2^{-2k}\right] = [0, 2^{-k}(1-2^{-k-1})].$$ At the next stage, we will remove from $I_k$ one interval of length $2^{-2(k+1)}$ from $I$, then two intervals of length $2^{-2(k+2)}$ and so on. So the total length of the intervals removed from $I_k$ is $$\sum_{n=1}^\infty 2^{n-1} 2^{-2(k+n)} = 2^{-2k} \sum_{n=1}^\infty 2^{-n-1} = 2^{-2k} \frac{1}{2} = 2^{-2k-1}.$$ Therefore, we have $$\frac{m(C \cap I_k)}{m(I_k)} = \frac{m(I_k) - 2^{-2k-1}}{m(I_k)} = \frac{2^{-k}(1-2^{-k-1}) - 2^{-2k-1}}{2^{-k}(1-2^{-k-1})} = \frac{(1-2^{-k-1}) - 2^{-k-1}}{(1-2^{-k-1})}$$ upon cancelling a factor of $2^{-k}$. It is clear by inspection that $$\lim_{k \to \infty} \frac{m(C \cap I_k)}{m(I_k)} = 1,$$ so given $\epsilon > 0$ you can choose $k$ so large that $m(C \cap I_k) > (1-\epsilon) m(I_k)$.
Best Answer
Here you find the Monotone convergence theorem from Lebesgue. In wikipedia is explained as integral of functions, but it can be stated also like that:
Let $A_1, \cdots, A_n, \cdots$ be a decreasing sequence of measurable sets, such that $m(A_1) < \infty$. Then (a) $A = \cap_{n=1}^{\infty} A_n$ is measurable, and (b) $m(A) = \lim_{n\to \infty}m( A_n) $.
The proof of (a) is quite simple from the basic definition, and (b) from monotony and $\sigma$-subaditivity.
A direct proof of $\mu ( C ) \geq \frac{1}{2} $ can be given, then, by $\sigma$-subaditivity of $\mu$ : As $C_1 \subseteq C \cup \bigcup_{n=1}^{\infty} (C_n \setminus C_{n+1})$, then $\frac{1 + 2^{-1}}{2} \leq \mu(C) + \sum_{n=1}^{\infty} \frac{2^{-n} - 2^{-n-1}}{2} = \mu (C) + \frac{ 2^{-1}}{2} $ if i'm not mistaken.