Need help proving $[0,1]$ is Borel isomorphic to $\{0,1\}^\mathbb{N}$. I am reading this note on probability theory
https://terrytao.wordpress.com/2015/10/12/275a-notes-2-product-measures-and-independence/
and in the proof of Lemma 12, professor Tao wrote that
"Let us call two topological spaces Borel isomorphic if their corresponding Borel structures are isomorphic as measurable spaces. Using the binary expansion, we see that $[0,1]$ is Borel isomorphic to $\{0,1\}^{\bf N}$ (the countable number of points that have two binary expansions can be easily permuted to obtain a genuine isomorphism)",
which I am having difficulty to justify. In other words, I need to find a bijection between $[0,1]$ and $\{0,1\}^\mathbb{N}$ such that both $f$ and $f^{-1}$ are measurable. In addition, I am not sure what topology to put on $\{0,1\}^\mathbb{N}$.
 A: Thank you all for your answers. I believe that I have justified this, although, I have not written it down rigorously yet. 
Here is the idea: let $f:[0,1] \to \{0,1\}^\mathbb{N}$ be the binary expansion of numbers in $[0,1]$. We choose $f(x)$ such that if $x$ has two different binary expansions, then choose $f(x)$ to be the one with infinitely repeating $0$ at the end. 
But $f$ is only injective. So we need to modify $f$ a little bit. Notice that the set of points $x$ that have two binary expansions is countable. In other words, the set of points in $\{0,1\}^\mathbb{N}$ that are not in $f([0,1])$ is countable. 
Put it differently, let $A$ be the subset of $[0,1]$ whose elements have two binary expansions. Then $A$ is countable, so we can write $A=\{a_n:n\in\mathbb{N}\}$. Originaly, we map each $a_n$ to $f(a_n)$, where $f(a_n)$ was chosen to be the expansion with infinitely repeating $0$ at the end. Call $f'(a_n)$ be the non-terminating expansion of $a_n$. Now, we modify $f$ such that $f$ maps $a_k$ to the set $\{f(a_n):n\in\mathbb{N}\}$ for $k$ odd, and $f$ maps $a_k$ to the set   $\{f'(a_n):n\in\mathbb{N}\}$ for $k$ even. Keep $f$ unchanged for other points. Then, $f$ is bijective.
Now, we need to show that $f$ and $f^{-1}$ is Borel measurable. This is easy:
To see why $f$ is measurable, take for example and open set in $\{0,1\}^\mathbb{N}$, to be something like  $E=\{1\}\times\{1\}\times\{0\}\times\{1\}\times\{0,1\}^\mathbb{N}$. Then $f^{-1}(E)$ will be a closed set such that maybe a countable points of $f^{-1}(E)$ are shifted. But that is not a problem because $f^{-1}(E)$ is still a Borel set.
For measurability of $f^{-1}$. Let $(a,b)$ be an open subset of $[0,1]$. Let $[a_n,b_n]$ be  such that $a_n, b_n$ have terminating binary expansion, and $[a_n,b_n]$ increases to $(a,b)$. It is easy to see that $f([a_n,b_n])$ is closed (mod a countable set, of course) and hence Borel. Thus, $f((a,b))$ is Borel.   
