Proof of Holder's Inequality using Jensen's Inequality - Integrability

Question

I did not find any posts regarding this: So I was reading the Alternate proof of Holder's inequality. What I am confused is the this bolden part:

From Jensen's inequality, $$ \int h \, d \nu = \Big( \int h^p \, d\nu\Big)^{\frac{1}{p}}$$ where $\nu$ is a probability measure, and $h$ is any $\nu$-measurable function.

The proof I know of requires $h$ to be integrable, or at least $\int h \,d \nu < \infty$.

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Supposing this condition is required. If we proceed the proof, we need to first show $$ \int hg \, d\nu < \infty$$ without invoking the inequality. But this seems to be what we are proving.

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I believe I am interpreting something wrong. What am I missing?

Answer 1

I think what's missing is that when $h$ is non-negative, if $\int h \,d\nu=\infty$, then $\int h^p\,d\nu=\infty$, so that Jensen's inequality applies to any $\nu$-measurable function.

Proof:

Let $A=h^{-1}([0,1))$. Let $B=A^C$. These are measurable since $h$ is. Then by assumption, $$\infty = \int h \,d\nu = \int_A h\,d\nu + \int_B h\, d\nu,$$ and $\int_Ah\,d\nu \le 1$ (since $h\le 1$ on $A$, and $\nu$ is a probability measure), so we must have $\int_Bh\,d\nu=\infty$. Then on $B$, $h\le h^p$, so $\int_Bh^P\,d\nu = \infty$, and then $\int h^p\,d\nu \ge \int_B h^p\,d\nu$. Hence $\int h^p\,d\nu=\infty$ as well.

Answer 2

Jensen's inequality, in this case, states that $$ \int h d\nu \leq \left(\int h^p d\nu\right)^{1/p} $$ provided that $\nu$ is a probability measure and $h$ is a non-negative $\nu$-measurable function. (There is no loss of generality assuming that $h\geq 0$, since we can replace a complex-valued $h$ with its module $|h|$.)

Clearly, both. sides of this inequality can be $+\infty$.

You do not need to first show that $\int fg d\nu$ is finite. Here $f$ and $g$ are again non-negative $\nu$-measurable functions, and that integral can be $+\infty$.