A question on klt pairs

If I interpreted correctly your question I guess the answer is positive. Here is the statement:

Let $f:Y\rightarrow X$ be a proper birational morphism between normal $\mathbb{Q}$-factorial varieties, and let $D$ be an effective $\mathbb{Q}$-divisor in $X$. Let us write $$K_Y = f^{*}(K_X+D)+\sum_{i}a_iE_i-\widetilde{D}$$ where $E_i$ are $f$-exceptional divisors and $\widetilde{D}$ is the strict transform of $D$. If $a_i \leq 0$ for any $i$, and $mult_y(\widetilde{D}-\sum_{i}a_iE_i) < 1$ for any $y\in Y$ , then the pair $(X,D)$ is klt.

To prove this you can proceed as follows. We have $\widetilde{D}-\sum_{i}a_iE_i = \widetilde{D}+\sum_{i}b_iE_i$ with $b_i = -a_i \geq 0$. Then, $\widetilde{D}+\sum_{i}b_iE_i$ is effective. Since $mult_y(\widetilde{D}+\sum_{i}b_iE_i) < 1$ for any $y\in Y$, by Proposition 9.5.13 of "Positivity in Algebraic Geometry 2" we have that the multiplier ideal $\mathcal{I}(\widetilde{D}+\sum_{i}b_iE_i)$ is trivial. Then, by Section 9.3.B of "Positivity in Algebraic Geometry 2" we get that $(Y,\widetilde{D}+\sum_{i}b_iE_i)$ is klt. Now, we may write $$K_Y+\widetilde{D}+\sum_{i}b_iE_i = K_Y+\widetilde{D}-\sum_{i}a_iE_i = f^{*}(K_X+D).$$ Finally, since $(Y,\widetilde{D}+\sum_{i}b_iE_i)$ is klt, by Lemma 3.10 in http://arxiv.org/abs/alg-geom/9601026 the pair $(X,D)$ is klt as well.