Absolute value bound of Lebesgue integral

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For the Riemann integral, we have the bound $$\left|\int_Af(x)dx\right|\leq\left(\sup_{x\in A}|f(x)|\right)\cdot\left|\int_Adx\right|$$

Do we have a similar bound for the Lebesgue integral, one like

$$\left|\int_Af(x)d\mu(x)\right|\leq\left(\sup_{x\in A}|f(x)|\right)\cdot\left|\int_Ad\mu(x)\right|?$$

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Yes, only $\sup_{x\in A}|f(x)|$ should be replaced by its essential supremum. The proof can be seen, for example, in Natanson, Chap. 5.