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This is clearly a measure on <math>A</math> with <math>\lambda(A)=1 \frac{}{}</math> | This is clearly a measure on <math>A</math> with <math>\lambda(A)=1 \frac{}{}</math> | ||
| − | Moreover, <math>\int_{A}fd\mu = \mu(A)\int_A f d\lambda \frac{}{}</math> | + | Moreover, <math>\int_{A}fd\mu = \mu(A)\int_A f d\lambda \frac{}{}</math>. |
| + | |||
| + | By Jensen's Inequality, we get | ||
| + | |||
| + | <math>\phi(\int_Afd\lambda) \leq \int_A\phi(f)d\lambda = \frac{\int_A\phi(f)d\mu}{\mu(A)}</math>. | ||
| + | |||
| + | So <math>\phi(\frac{\int_Afd\mu}{\mu(A)}) \leq \frac{\int_A\phi(f)d\mu}{\mu(A)}</math> | ||
Latest revision as of 09:52, 22 July 2008
Define a function from the set of all measurable subset $ B $ of $ A $ as below
$ \lambda(B)=\frac{\mu(B)}{\mu(A)} $
This is clearly a measure on $ A $ with $ \lambda(A)=1 \frac{}{} $
Moreover, $ \int_{A}fd\mu = \mu(A)\int_A f d\lambda \frac{}{} $.
By Jensen's Inequality, we get
$ \phi(\int_Afd\lambda) \leq \int_A\phi(f)d\lambda = \frac{\int_A\phi(f)d\mu}{\mu(A)} $.
So $ \phi(\frac{\int_Afd\mu}{\mu(A)}) \leq \frac{\int_A\phi(f)d\mu}{\mu(A)} $
