Let X be a random variable with mean $\mu$ and continuous cumulative distribution function $F$. Show that
$$\int_{-\infty}^{a}F(x)dx = \int_{a}^{\infty}[1-F(x)]dx$$
if and only if $a = \mu$.
This question seems related, and I believe that changing the order of integration is involved but I haven't been able to work out the details.
Integrate by parts. \begin{equation*} \int_{-\infty}^{a}x f(x) dx =a F(a)-\int_{-\infty}^{a}F(x)dx \end{equation*} \begin{equation*} \int_{a}^{\infty}x f(x) dx=a(1-F(a))+\int_{a}^{\infty}(1-F(x))dx \end{equation*} Therefore \begin{equation*} \mu =\int_{-\infty}^{\infty}x f(x) dx= a-\int_{-\infty}^{a}F(x)dx+\int_{a}^{\infty}(1-F(x))dx \end{equation*}
Similar to the other question, you can write
$$\begin{align*} E[X] &= E[X \cdot I\{X > 0\}] + E[X \cdot I\{X \le 0\}] \\ &= E\Big[\int_0^\infty I\{X > t\} \, dt\Big] - E\Big[\int_{-\infty}^0 I\{X \le t\}\Big] \\ &= \int_0^\infty 1 - E[I\{X \le t\}] \, dt - \int_{-\infty}^0 E[I\{X \le t\}] \, dt \\ &= \int_0^\infty 1 - F(t) \, dt - \int_{-\infty}^0 F(t) \, dt \\ &= \int_0^a 1 - F(t) \, dt + \int_a^\infty 1 - F(t) \, dt - \int_{-\infty}^a F(t) \, dt - \int_a^0 F(t) \, dt \\ &= a + \int_a^\infty 1 - F(t) \, dt - \int_{-\infty}^a F(t) \, dt\,. \end{align*}$$
