Let $g$ be a non-decreasing function defined on the real line and bounded above by 1:
Assume $g(0)\ge 0$. Prove that $P(X<0)\le 1-E[g(X)-g(0)]$
Attempt:
Consider: $1-E[g(X)-g(0)]=1-E[g(X)-g(0)(X \ge 0)]-E[g(X)-g(0)(X < 0)]$
Since $E[g(X)-g(0)(X < 0)]<0$ then $-E[g(X)-g(0)(X < 0)]>0$
Then $1-E[g(X)-g(0)(X \ge 0)]-E[g(X)-g(0)(X < 0)]> 1-E[g(X)-g(0)(X \ge 0)]$
Note since $1\ge g(X)-g(0)$, then $1-E[g(X)-g(0)(X \ge 0)\ge 1-E[X\ge0]=P(X<0)$.
Concluding, $P(X<0)\le 1-E[g(X)-g(0)]$
Is my logic correct? The question asks also to find an example where $g(0)<0$ then $P(X<0)> 1-E[g(X)-g(0)]$ Have not been able to find an example can someone provide one?