Probability with Martingales:
How exactly do we have the part in the $\color{red}{\text{red}}$ box?
What I tried:
$$E\left[ \sum_{n=1}^{\infty} 1_{|X| > n} \right]$$
$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1_{|X| > n} + \sum_{n=\lfloor|X|\rfloor}^{\infty} 1_{\lfloor|X|\rfloor > n} \right]$$
$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1_{\Omega} + \sum_{n=\lfloor|X|\rfloor}^{\infty} 1_{\emptyset} \right]$$
$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1 + \sum_{n=\lfloor|X|\rfloor}^{\infty} 0 \right]$$
$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1 \right]$$
$$ = E\left[ \sum_{n=\color{blue}{1}}^{\color{red}{\lfloor|X|\rfloor - 1}} \color{green}{1} \right]$$
$$ = E\left[ (\color{red}{\lfloor|X|\rfloor - 1} - \color{blue}{1} + 1) (\color{green}{1}) \right]$$
$$ = E\left[ (\color{red}{\lfloor|X|\rfloor - 1}) (\color{green}{1}) \right]$$
$$ = E\left[ \color{red}{\lfloor|X|\rfloor - 1} \right]$$
$$ = E\left[ \lfloor|X|\rfloor - 1 \right]$$
$$ \le E\left[ |X| \right]$$
Is that right?
$$E\left[ \sum_{n=1}^{\infty} 1_{|X| > n} \right]$$
$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1_{|X| > n} + \sum_{n=\lfloor|X|\rfloor}^{\infty} 1_{\lfloor|X|\rfloor > n} \right]$$
$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1_{\Omega} + \sum_{n=\lfloor|X|\rfloor}^{\infty} 1_{\emptyset} \right]$$
$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1 + \sum_{n=\lfloor|X|\rfloor}^{\infty} 0 \right]$$
$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1 \right]$$
$$ = E\left[ \sum_{n=\color{blue}{1}}^{\color{red}{\lfloor|X|\rfloor - 1}} \color{green}{1} \right]$$
$$ = E\left[ (\color{red}{\lfloor|X|\rfloor - 1} - \color{blue}{1} + 1) (\color{green}{1}) \right]$$
$$ = E\left[ (\color{red}{\lfloor|X|\rfloor - 1}) (\color{green}{1}) \right]$$
$$ = E\left[ \color{red}{\lfloor|X|\rfloor - 1} \right]$$
$$ = E\left[ \lfloor|X|\rfloor - 1 \right]$$
$$ \le E\left[ |X| \right]$$