How does finite integral implies convergence almost everywhere in this proof in Stein-Shakarchi's Functional Analysis?

328 Views Asked by At

I am trying to understand a step in the following proof of completeness of $L^p$ in Stein-Shakarchi's Functional Analysis. (See the proof on page 5 of the link or at the end of this post.)

In the proof, it is said that:

... applying the monotone convergence theorem implies that $\int g^p<\infty$, and therefore, the series defining $g$ and hence the series defining $f$ converges almost everywhere, and $f\in L^p$.

I do not understand why.

Question 1: How does one get from the above argument that the almost everywhere convergence?

Question 2: How does one get that $f\in L^p$?


enter image description here enter image description here enter image description here

1

There are 1 best solutions below

13
On BEST ANSWER

Since $g \geq 0$ and $\int g^p<\infty$, $g^p$ (and thus $g$) is finite almost everywhere. Thus the series defining $g$ converges almost everywhere.

Since $|S_K(f)(x)| \leq S_K(g)(x)$ for all $x$, the series defining $f(x)$ converges for almost every $x$.

Finally, $|f| \leq |g|=g$, so $\int|f|^p \leq \int g^p < \infty$, so $f \in L^p$.