Markov inequality with $>$ in place of $\geq$ The usual Markov inequality goes as follows: suppose $X$ is a random variable with $g \geq 0$. Then for any $r &gt; 0$, 
$$\mathbb{P}(g(X) \geq r) \leq \dfrac{\mathbb{E}[g(X)]}{r}\text{.}$$
Does the following statement still hold true?

$$\mathbb{P}(g(X) &gt; r) &lt; \dfrac{\mathbb{E}[g(X)]}{r}\text{.}$$

For context, suppose $\mathbb{E}[|X|] = 0$ and we want to find $\mathbb{P}(X &gt; 1)$. A solution I read said that since $\mathbb{P}(X &gt; 1) \leq \mathbb{P}(|X| &gt; 1)$, we can show $\mathbb{P}(X &gt; 1) = 0$ using the above.

I understand how the Markov inequality led to $\mathbb{P}(X &gt; 1) = 0$, but I'm not comfortable using it in the $&gt;$ case without justification.
The strict inequality will not be true in general, because it could be that $\mathbb{P}(g(X)&gt;r)=\mathbb{P}(g(X)\geq r)=\frac{\mathbb{E}[g(X)]}{r}$.

But you can say that $\mathbb{P}(g(X)&gt;r)\leq \mathbb{P}(g(X)\geq r)\leq \frac{\mathbb{E}[g(X)]}{r}$, which should be enough for most applications.
It does hold true iff $g(X) = 0$ does not hold almost surely. Here is a short proof:

$$
  \mathbb{E}[g(X)] \\= \mathbb{P}(g(X) \leq r) \mathbb{E}[g(X)|g(X) \leq r] + \mathbb{P}(g(X) &gt; r) \mathbb{E}[g(X)|g(X) &gt; r] \\&gt; g(r) \mathbb{P}(g(X) &gt; r)
$$

The last inequality holds since either $\mathbb{P}(g(X) &gt; r) &gt; 0$ or the first term (and the whole expression) will simply be $\mathbb{E}[g(X)] &gt; 0$. Dividing by $g(r)$ yields $\mathbb{P}(g(X) &gt; r) &lt; \frac{\mathbb{E}[g(X)]}{g(r)}$.