proof an inequality; measure theory

**Rapha** · March 29th 2010, 08:29 AM

Hi there

Let C be compakt
$\mu$ a radon measure and $\mu(C) < \infty$ .

Let $v \in L^1(\mathbb{R}) , \ 0 < a < b < \infty$

Do you know how to proof the following inequality:

$\int_{a}^b |v| d\mu \le \mu([a,b]) \cdot \|v\|_\infty$

I don't get it.
By the way I'm not sure about the correctness of this inequaility if $\mu$ is an arbitrary measure. If it's correct where $\mu$ equals the Lebesgue measure, you can use this instead of the Radon measure.

Any comments are welcome

Yours Rapha

**Opalg** · March 29th 2010, 12:37 PM

Presumably $\mu$ is supposed to be a positive Radon measure.

From the definition of the $L^\infty$ -norm, $|v(t)|\leqslant \|v\|_\infty$ for almost all t in [a,b]. Then $\int_a^b\!\!\!|v(t)|\,d\mu(t) \leqslant \int_a^b\!\!\!\|v\|_\infty\,d\mu(t) = \mu([a,b]) \|v\|_\infty$ (integral of a constant is the constant times the measure of the interval).

**Rapha** · March 30th 2010, 01:14 AM

Hello Opalg

thank you very much!

Kind regards
Rapha

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