If $E(X^-)=0$ then $P(X\geqslant0)=1$

Question

Let $I(t)$ a real-valued stochastic process with continuous trajectories. Let $I^-(t)$ the negative part of $I(t)$, so $I(t)=I^+(t)-I^-(t)$. I know that $\mathbb{E}[I^-(t)]=0$, and I have to prove that $\mathbb{P}(I(t) \ge0)=1$.

I tried to argue by absurd to prove $\mathbb{P}(I(t) <0)=0$ but I'm not able to reach a contraddiction.