Prove that $\int_E \log fd\mu \leqslant \mu(E) \, \log \left[\frac{1}{\mu(E)} \right]$ for strictly positive measure $\mu$

Question

Let $(X, \Sigma, \mu)$ be a measurable space, with strictly positive measure $\mu$.

Let $f: X \rightarrow (0, \infty)$ be a function such that $\int_X fd\mu=1$.

Show that for every $\mu$-measurable set $E \in \Sigma$ with $0 < \mu(E)<\infty$ the following inequality holds:

$$\int_E \log fd\mu \leqslant \mu(E) \, \log \left[\frac{1}{\mu(E)} \right]$$

I'm dealing with this quiestion quite a bit and can't seems to find a solution. My approach: We define new measure $\nu = f \cdot \mu$. Since $f>0$ and $\mu$ strictly positive, then $\nu$ is strictly positive.

Moreover, $\nu(X) \equiv \int_Xdv = \int_Xd(f\cdot\mu) =\int_Xfd\mu = 1$. Therefore, by the Jansen inequality (concaved version):

For every integrable function $g$ (according to $\mu$ ) such that $g:X \rightarrow (0,\infty)$ and concave function $F:(0, \infty) \rightarrow$ R (R for the real numbers):

$$\int_X F\circ g d\nu \leq F \biggl(\int_X g d\nu \biggl)$$

I thought about taking $F(x) = \log(x)$ but I can't seem to find the right $g$ and how to conclude it to the integral overall given set $E$.

Does anyone have a solution for this? Thanks.

Edit: I got it. Take $g=f$ and for every set E above define $\nu_E = \frac{1}{\mu(E)} \cdot \mu$. Clearly, $\nu(E)=1$ there fore by the Jensen inequality we get:

$$\int_E \frac{1}{\mu(E)} \cdot \log f d\mu \leq \log \biggl(\int_E f \frac{1}{\mu(E)} \cdot d\mu \biggl)$$

Now we play with the last inequality and we get an equivalent one: $$\int_E \log fd\mu \leqslant \mu(E) \, \log \left[\frac{1}{\mu(E)} \right]$$

Answer 1

The measure $\frac{1}{\mu(E)} \mu$ is a probability measure on $E$. Since the exponential function is convex, Jensen's inequality gives you $$ \exp \left[ \frac{1}{\mu(E)} \int_E \log f \, d\mu \right] \le \frac{1}{\mu(E)} \int_E \exp [\log f] \, d\mu = \frac{1}{\mu(E)} \int_E f \, d\mu \le \frac{1}{\mu(E)}.$$ Now take the logarithm on both sides to find $$ \frac{1}{\mu(E)} \int_E \log f \, d\mu \le \log \left( \frac{1}{\mu(E)} \right).$$

Answer 2

Define

$$\mu_E(dy) := \frac{1}{\mu(E)} 1_E(y) \, d\mu(y)$$

then

$$\int_E \log f \, d\mu = \mu(E) \int \log f \, d\mu_E.$$

Applying Jensen's inequality to the probability measure $\mu_E$ and $F(x) = \log x$ we find

$$\int_E \log f \, d\mu \leq \mu(E) \log \left( \int f \, d\mu_E \right) = \mu(E) \log \left( \frac{1}{\mu(E)} \int_E f \, d\mu \right).$$

As $\int_E f \, d\mu \leq 1$ it follows from the monotonicity of the logarithm that

$$\int_E \log f \, d\mu \leq \mu(E) \log \left( \frac{1}{\mu(E)} \right). $$

Answer 3

I think you had an almost correct idea. For fixed $E$, we can rescale $d\nu = \frac{d\mu}{\mu(E)}$ so that $ (E,\nu)$ is a probability space. Then Jensen's applies to the concave function $\log$ to give $$ \int_E \log f d\nu ≤ \log\left( \int_E f d\nu\right)$$ which rewrites to $$ \frac{1}{\mu (E)}\int_E \log f d\mu ≤ \log\left( \frac{1}{\mu (E)} \int_E f d\mu\right) ≤ \log \left(\frac{1}{\mu (E)}\right)$$

which is the result.