I am reading the book Theoretical Statistics: Topics for a Core Course from Keener, 2010. In section 8.4 Medians and Percentiles includes:
"Let $X_{1},...,X_{n}$ be random variables. These variables, arranged in increasing order, $X_{(1)}\le X_{(2)}\le\cdots\le X_{(n)}$, are called order statistics.
The median is the middle order statistic:
$ \tilde{X}=\begin{cases} X_{(m)},&n=2m-1\\ \frac{1}{2}(X_{(m)}+X_{(m+1)}),&n=2m \end{cases} $
Assume now that $X_{1},...,X_{n}$ are iid with common cumulative distribution function $F$, and let $\tilde{X}_{n}$ be the median of the first $n$ observations. For regularity, assume that $F$ has a unique median $\theta$, so $F(\theta)=1/2$, and that $F'(\theta)$ exists and is finite and positive. Let us try to approximate
$P\left(\sqrt{n}(\tilde{X}_{n}-\theta)\le a\right)=P\left(\tilde{X}_{n}\le\theta+(a\ /\sqrt{n})\right)$
Define $S_{n}=\sum_{i=1}^{n}\mathbb{1}\{X_{i}\le\theta+(a\ /\sqrt{n})\}$.
The key to this derivation is the observation that $\tilde{X}_{n}\le\theta+(a\ /\sqrt{n})$ if and only if $S_{n}\ge m$."
My question is how to prove the statement "$\tilde{X}_{n}\le\theta+(a\ /\sqrt{n})$ if and only if $S_{n}\ge m$"?
For simplicity, let $n$ be odd.
$\Rightarrow$: If $S_n \geq m$, then at least $m$ of your variables $X_i$ satisfy $X_i \leq \theta + a / \sqrt{n}$. Without loss of generality, suppose these variables are $X_1, X_2, \dots, X_m$. Then it's easy to see that by definition, $\tilde{X}_n \leq \max_{1 \leq j \leq m} X_{j} \leq \theta + a / \sqrt{n}$.
$\Leftarrow$: If the median satisfies $\tilde{X}_n \leq \theta + a / \sqrt{n}$, then $X_{(1)}, \dots, X_{(m)}$ must all satisfy the same inequality as well. But that means that at least $m$ indicators of the form $\mathbf{1}\{X_i \leq \theta + a / \sqrt{n}\}$ will evaluate to $1$, and thus $S_n \geq m$.