Outer Measure on $\mathbb{R}$ Let $p\ge0$, $d>0$ and $X \subset \mathbb{R}$.
(i) Define:
$$H_{p,d}(X) := \inf\left\{\sum_{i=1}^{\infty}\operatorname{diam} \left(\frac{A_i}{2}\right)^p : X \subset  \bigcup_{i=1}^{\infty}A_i, 0 < \operatorname{diam}A_i  \leq d \right\}.$$
If $X$ is empty, $H_{p,d}(X)=0$ .
(ii) Now, define
$$H_p(X) := \lim_{d\rightarrow 0} H_{p,d}(X).$$

Prove that $H_p$ is an outer measure on $\mathbb{R}$ .

I have proved that $H_{p,d}$ is an outer measure on $\mathbb{R}$. Using that I was trying to show that $H_p$ is an outer measure on $\mathbb{R}$ , but I cannot prove the countable subadditivity of $H_p$.
 A: You already proved that $H_{p,d}$ is an outer measure, so for a sequence of subsets $X_i$ of $\mathbb{R}$, we have
$$ H_{p,d}\left(\bigcup_{i\in\mathbb{N}} X_i\right)\leq \sum_{i=1}^{\infty}H_{p,d}(X_i).$$
Taking limits yields
$$ H_{p}\left(\bigcup_{i\in\mathbb{N}} X_i\right)\leq \lim_{d\rightarrow 0}\lim_{n\rightarrow\infty}\sum_{i=1}^{n}H_{p,d}(X_i), $$ 
so showing that the two limits on the right can be interchanged would show the countable subadditivity. 
(1) We may assume that
$$ L:=\sum_{i=1}^{\infty}H_{p}(X_i) < \infty $$
as otherwise the required inequality holds trivially. 
(2) Note that $H_{p,d}$ is decreasing as a function of $d$ as the set 
$$\left\{A\supset X: 0< \operatorname{diam} A \leq d \right\} $$
grows as $d$ grows, so the infimum in the definition of $H_{p,d}$ can only get smaller.
(3) Next, consider the sequence of functions defined by
$$ f_n(d) := \sum_{i=1}^n H_{p,d}(X_i). $$
By monotony, and since $L$ is required to be finite, the limit
$$ H_{p,0}(X_i) := \lim_{d\rightarrow 0} H_{p,d}(X_i)$$
exists and is the maximum value of $H_{p,d}$. We thus observe that
$$ \sum_{i=1}^n \left\|H_{p,d}(X_i)\right\|_{\infty} = \sum_{i=1}^n H_{p,0}(X_i) = \sum_{i=1}^n H_{p}(X_i) \leq L <\infty $$
for all $n\in\mathbb{N}$, where $\left\|\cdot\right\|_{\infty}$ denotes the supremum norm, i.e. the series
$$\sum_{i=1}^{\infty} \left\|H_{p,d}(X_i)\right\|_{\infty}$$
converges. This shows that the sequence $f_n$ is uniformly convergent with respect to $d$, hence the limit function $f$ is continuous (at least) at $d=0$.
Using continuity, we finally get
$$ \lim_{d\rightarrow 0}\lim_{n\rightarrow\infty}\sum_{i=1}^{n}H_{p,d}(X_i) = \lim_{d\rightarrow 0 } f(d) = f(0) = \lim_{n\rightarrow\infty}f_n(0) = \sum_{i=1}^{\infty}H_{p}(X_i)$$
as was to be shown.
