\rceil$ and the reduced divisor $C  := (F_{\rm 0})_{\rm red} $ on $X$, the above exact sequence implies that $\chi({{\mathcal{O}}}_C(M|C) ) = \chi({{\mathcal{O}}}_X(M)) - \chi({{\mathcal{O}}}_X(M-C))
= C . M - C . (K_X + C)/2$, where we applied the Riemann-Roch theorem for both ${{\mathcal{O}}}_X(M)$ and ${{\mathcal{O}}}_X(M - C)$. Now $C . (K_X + C) = 0$ and $C . M \ge (F_{\rm 0})_{\rm red} . (K_X + aL) > 0$ (for $0 \ne C$ being nef and $K_X + aL$ nef and big), so $\chi({{\mathcal{O}}}_C(M|C)) > 0$. By the vanishing above, $h^0(X, K_X + \lceil aL \rceil) = \chi({{\mathcal{O}}}_X(M))  = \chi({{\mathcal{O}}}_X(M-C)) + \chi({{\mathcal{O}}}_C(M|C) ) =
h^0(X, K_X + \lceil aL \rceil - (F_{\rm 0})_{\rm red}) ) + \chi({{\mathcal{O}}}_C(M|C) )  > 0  + 0$. This proves the theorem.

The above argument actually proved the following: let $\pi : X
\rightarrow B$ be a relatively minimal elliptic surface with $\kappa(X) = -\infty$. Suppose that $L$ is a nef and big ${\bf Q}$-divisor such that $K_X + L$ is nef. Then $H^0(X, K_X + \lceil aL \rceil) \ne 0$ provided that either $a > 1$, or $a = 1$ and $K_X + L$ is nef and big.

Preparations for surfaces with $\kappa = 