I am trying to prove that:
Theorem. Let $F$ be a Borel subset of $\mathbb{R}^{n}$. Then $\mathcal{H}^{n}(F)=c_{n}^{-1}\text{vol}^{n}(F)$ where $c_{n}$ is the volume of the $n$-dimensional ball of diameter $1$ and $\mathcal{H}^{n}$ is the $n$-dimensional Hausdorff measure.
Proof (so far). Let $F$ be a Borel subset of $\mathbb{R}^{n}$, then $$\mathcal{H}^{n}(F)=\lim_{\delta\to0}\left(\inf\left\{\sum_{i=1}^{\infty}|U_{i}|^{n}:\{U_{i}\} \text{ is a } \delta\text{-cover of } F\right\}\right)$$
I am not sure how to then show that the RHS is $c_{n}^{-1}\text{vol}^{n}(F)$. Can someone help me please?
By definition, the Lebesgue measure (or volume) of a Borel set $F \subseteq \mathbb{R}^n$ is $$ m(F) := \inf\left\{ \sum_{j=1}^{\infty} m(B_j) : \text{$\{B_j\}$ is a collection of open balls such that $F \subseteq \bigcup_{j=1}^{\infty} B_j$}\right\}. $$
Note that the measure of a ball in $\mathbb{R}^n$ is constant multiple of the radius to the $n$-th power. That is, if $B(x,r)$ is the open ball centered at $x$ with radius $r$, then $$ m(B(x,r)) = c_j r^n, $$ where $c_n$ is the measure of a ball of radius 1. For example, the measure of a ball in $\mathbb{R}^2$ is $\pi r^2$, and the measure of a ball in $\mathbb{R}^3$ is $\frac{4}{3} \pi r^3$ (these formulae should, I hope, ring a bell). This implies that $$ \sum_{j=1}^{\infty} m(B_j) = \sum_{j=1}^{\infty} c_j r_j^n, $$ where $r_j$ is the radius of the $j$-th ball in the collection. Observe that if $\{U_j\}$ is a $\delta$-cover of $F$, then for each $j$ we can find a ball $B_j = B(x_j, r_j)$ such that $U_j \subseteq B(x_j,r_j)$ and $r_j = |U_j|/2$. That is, every $\delta$-cover corresponds to a $\delta$-cover by open balls. Hence we have \begin{align} \mathcal{H}_\delta^n(F) &= \inf\left\{ \sum_{j=1}^{\infty} |U_j|^n : \text{$\{U_j\}$ is a $\delta$-cover of $F$} \right\} \\ &= \inf\left\{ \sum_{j=1}^{\infty} |B_j|^n : \text{$\{B_j:=B(x_j,r_j)\}$ covers $F$, $r_j < \frac{\delta}{2}$} \right\} \\ &\ge \inf\left\{ \sum_{j=1}^{\infty} r_j^n : \text{$\{B_j:=B(x_j,r_j)\}$ covers $F$} \right\} \tag{$\ast$} \\ &= \inf\left\{ c_n^{-1} \sum_{j=1}^{\infty} c_n r_j^n : \text{$\{B_j:=B(x_j,r_j)\}$ covers $F$} \right\} \\ &= c_n^{-1} \inf\left\{ \sum_{j=1}^{\infty} m(B_j) : \text{$\{B_j:=B(x_j,r_j)\}$ covers $F$} \right\} \\ &= c_n^{-1} m(F), \end{align} where the inequality at $(\ast)$ follows from the fact that if $X\subseteq Y \subseteq \mathbb{R}$, then $\inf(X) \ge \inf(Y)$ (the intuition is that the smallest thing in $Y$ could be the smallest thing in $X$, or it could be even smaller, as there are more things in $Y$). Taking $\delta\to 0$, this implies that $$ \mathcal{H}^n(F) = \lim_{\delta\to 0} \mathcal{H}^n_{\delta}(F) \ge c_n^{-1} m(F), $$ which is half of what we need to show.
The other inequality is (perhaps) a bit tricker. First, note that if $m(F) = \infty$, we are done, so assume that $m(F) < \infty$. I am now going to rely on some basic results about Lebesgue measure in $\mathbb{R}^n$, without proving them. Fix some $\varepsilon > 0$ and $\delta > 0$. There exists a collection of balls $\{B_j = B(x_j, r_j)\}$ such that
In particular, note that $\{B_j\}$ is a $\delta$-cover of $F$. But then we have \begin{align} m(F) + \varepsilon &> \sum_{j=1}^{\infty} m(B_j) \tag{by (3), above} \\ &= c_n \sum_{j=1}^{\infty} r_j^n \\ &\ge c_n \inf\left\{ \sum_{j=1}^{n} |U_j|^n : \text{$\{U_j\}$ is a $\delta$-cover of $F$} \right\} \\ &= c_n \mathcal{H}^{n}_{\delta}(F). \end{align} Since $\varepsilon$ and $\delta$ were arbitrary, it follows that we can take limits in order to obtain \begin{align} \lim_{\varepsilon\to 0} \left(\lim_{\delta\to 0} m(F)\right) \ge \lim_{\varepsilon\to 0} \left(\lim_{\delta\to 0} c_n \mathcal{H}^{n}_{\delta}(F)\right) &\implies \lim_{\varepsilon\to 0} \left( m(F) + \varepsilon \right) \ge c_n \lim_{\varepsilon\to 0} \mathcal{H}^n(F) \\ &\implies m(F) \ge c_n \mathcal{H}^n(F). \end{align} This gives the opposite inequality, and finishes the proof.