What if a random variable is (surely/almost surely) greater than (or even less than) its mean? $X \ge \mu$ ($X \le E[X]$) Assume that $X$ is integrable in probability space $(\Omega, \mathscr F, \mathbb P)$, i.e. $X \in \mathscr L^1 (\Omega, \mathscr F, \mathbb P)$.

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*What does it mean if a random variable is (surely/almost surely) greater than its expected value? $X \ge E[X]$ I think this means $X$ is, at least almost surely, constant. (I'm not sure $X$ is surely constant even if $X \ge E[X]$ surely.) How do we prove this though? What I've done so far:

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*1.1. I can prove this for $X$ indicator and nonnegative simple (and nonnegative discrete).

*1.2. I didn't bother anymore to try for nonnegative integrable and general integrable because I'm hoping for some simple proof I might've over looked like...

*1.3. ...like prove that $P(X > E[X]) = 0$ through something like this. Maybe consider $E[X1_A]$ where $A=\{X > E[X]\}$ or something.

*1.4. If standard machine is really the way to go about this, then I'm stuck: For nonnegative integrable, probably monotone convergence theorem, but not really sure how. But since we're still in nonnegative, I'm guessing we'll have $X=0$. For general integrable, ok this part I remember is actually not just simple but also easy, so I must really be over looking something.



*Does the same conclusion in (1) (I mean whatever is the correct conclusion and not necessarily what I have stated) hold if $X$ is instead (surely/almost surely) less than its expected value? $X \le E[X]$


*Elementary/basic probability theory: If $X$ is a continuous random variable, then how do we show it is impossible that $X \ge E[X]$ surely (and also $X \le E[X]$ surely) (and also almost surely, but you know, it's still elementary/basic)? (I guess ignore this part if you can answer the above without measure theory.)


*If answering any of the above is easier if we assume $X$ is square integrable, then please tell me how (eg somehow we can say $Var[X]=0$).
 A: This can be shown through measure theory. Let's use your $A=\{X>\mu\}$. Then on $A$, $X-\mu$ is positive. On its complement $A^c$, $X$ equals $\mu$ so $X-\mu$ is zero.
Hence:
$$\mu=E[X]= \mu+ \int_A X-\mu dP + \int_{A^c} X-\mu dP = \mu + \int_A X-\mu dP.$$
Now note that the final term has to be zero and the integrand is strictly positive on $A$. Hence, $A$ has to be a null-set.
You can try to follow a similar proof strategy for the case of a continuous random variable with a pdf.
A: We show that

If $X\in L_1(\Omega,\mathscr{F},\mathbb{P})$ and $X\geq \mathbb{E}[X]$ $\mathbb{P}$-a.s. then $X=\mathbb{E}[X]$ $\mathbb{P}$-a.s.

This will be a consequence of the following observation:

Suppose $\mathbb{P}[A]>0$ and $g$ is a measurable function that strictly positive on $A$, then $\mathbb{E}[g\mathbb{1}_A]=\int_A g\,d\mathbb{P}>0$.

To see this, notice that $\{\omega\in A:g(\omega)>0\}=\bigcup^\infty_{n=1}\{\omega\in A: g(\omega)>\frac1n\}$. Since $\mathbb{P}\big[\{\omega\in A: g(\omega)>0\}\big]>0$, there is $n_0\in\mathbb{N}$ such that $\mathbb{P}\big[\{\omega\in A: g(\omega)>\frac{1}{n_0}\}\big]>0$. Then, by Markov-Chebyshev's inequality
$$\int_Ag\,d\mathbb{P}\geq \int_{A\cap\{g>\tfrac{1}{n_0}}g\,d\mathbb{P}\geq \frac{1}{n_0}\mathbb{P}\big[A\cap\{g>\tfrac{1}{n_0}\}\big]>0\qquad\Box.$$
To conclude the  suppose $A=\{X> \mathbb{E}[X]\}$ has positive measure.
Define $g(\omega)=(X-\mathbb{E}[X])\mathbb{1}_A(\omega)$. Notice that $\{g>0\}=A$ and by assumption, $\mathbb{P}[A]>0$. Then
$$0<\int_Ag\,d\mathbb{P}=\int_A(X-\mathbb{E}[X])\,d\mathbb{P}\leq \int_\Omega(X-E[X])\,d\mathbb{P}=0$$
which yields a contradiction. Hence, $\mathbb{P}[A]=0$ and so, $X\leq \mathbb{E}[X]$ $\mathbb{P}$-a.s. This, along with the assumption that $X\geq\mathbb{E}[X]$ $\mathbb{P}$-a.s., implies that $X=\mathbb{E}[X]$ $\mathbb{P}$-a.s.


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*A similar conclusion follows if one assumes that $X\leq \mathbb{E}[X]$ $\mathbb{P}$-a.s.

