Show that Lebesgue $\sigma$-algebra has the same cardinality as $\mathcal{P}(\Bbb R)$ We recently started with measure theory in my Analysis class and I got stuck trying to solve the following exercise:
Show that Lebesgue $\sigma$-algebra on $\Bbb R$ has the same cardinality as $\mathcal{P}(\Bbb R)$. There is also a tip, suggesting that I should have a look at subsets of $\mathcal{C}$ (standard Cantor set on unit interval).
Using the tip I believe I should look for bijection between $\mathcal{C}$ and $\mathcal{P}(\Bbb R)$. Since both of those sets are uncountable, how can I come up with such bijection? Also to be honest, I do not fully get the argument why is it sufficient to search for bijection between $\mathcal{C}$ and $\mathcal{P}(\Bbb R)$ instead of bijection between Lebesgue $\sigma$-algebra on $\Bbb R$ and $\mathcal{P}(\Bbb R)$.
Any tip\solution would be greatly appreciated. Thanks for the help in advance!
 A: Two important facts about the Cantor middle-thirds set:


*

*It has Lebesgue measure zero.  Indeed, the $n$th Cantor iterate has measure $(\frac{2}{3})^n$, so the full Cantor set has at most that.

*It has the same cardinality as $\mathbb{R}$.  One outline for why:  Points in the Cantor set have ternary expansions (that is, base-3 "decimal" expansions) consisting solely of 0's and 2's.  So, the cardinality of the Cantor set is the same as the number of sequences of 0's and 2's.  This is, of course, the same as the number of sequences of 0's and 1's.  Sequences of the latter form can be identified with points in the unit interval [0,1] by interpreting them as binary expansions (i.e. base 2 "decimals").  Finally, the interval [0,1] is the same cardinality as $\mathbb{R}$; it has the same cardinality as (0,1), and the latter has explicit bijections with $\mathbb{R}$ such as $x \mapsto \tan(\pi(x-1/2))$.


Other prerequisites: 


*

*Every subset of a Lebesgue-measure-zero set is Lebesgue measurable.  

*If two sets have the same cardinality, their power sets have the same cardinality.  (A bijection between the sets can be extended in a natural way to a bijection between their powersets.)


Now let $\mathfrak{L}$ be the class of Lebesgue-measurable sets, $C$ the Cantor set.  We know
$$
P(\mathbb{R}) \simeq P(C) \subseteq \mathfrak{L}
$$
which implies that the cardinality of $\mathfrak{L}$ is at least as great as that of $P(\mathbb{R})$.  On the other hand, $\mathfrak{L}$ is a subset of $P(\mathbb{R})$, so its cardinality cannot be strictly greater.  Hence $\mathfrak{L}$ and $P(\mathbb{R})$ have equal cardinality.  (Cantor-Bernstein is in the background for these last few sentences.)
