integration – Mathtuition88

Necessary and Sufficient Condition for Integrability in finite measure space

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Let $(\Omega, \Sigma, \mu)$ be a finite measure space. Suppose that $f\geq 0$ is a measurable function on $\Omega$ . Let $E_n:=\{\omega\in\Omega:f(\omega)\geq n\}$ for each $n\in \mathbb{N}\cup\{0\}$ . Show that $f$ is integrable if and only if $\sum_{n=0}^\infty \mu(E_n)<\infty$ .

This proof has a cute solution that is potentially very short. We will elaborate more on this proof. Other approaches include using Markov’s Inequality / Chebyshev’s inequality.

Proof: Consider $g=\sum_{n=1}^\infty I_{E_n}$ .

Note that for each $n\geq 0, n\in\mathbb{N}$ on $E_n\setminus E_{n+1}$ , $g=n$ , while $n\leq f<n+1$ . Therefore $g\leq f<g+1$ on $\Omega$ .

Integrating with respect to $\mu$ , we get $\int_{\Omega} gd\mu\leq\int_{\Omega} fd\mu<\int_{\Omega} g+1d\mu$ .

$\displaystyle\boxed{\sum_{n=1}^\infty \mu (E_n)\leq \int_\Omega f d\mu<\sum_{n=1}^\infty\mu(E_n)+\mu(\Omega)}$

(=>) Now assuming f is integrable, i.e. $\int fd\mu<\infty$ , we have $\sum_{n=1}^\infty \mu(E_n)<\infty$ . $\mu(E_0)=\mu(\Omega)<\infty$ . Therefore $\sum_{n=0}^\infty\mu(E_n)<\infty$ .

(<=) Conversely, if $\sum_{n=0}^\infty\mu(E_n)=\sum_{n=1}^\infty \mu(E_n)+\mu(\Omega)<\infty$ , then $\int_\Omega f d\mu<\infty$ .

We are done.

Note: For a more rigorous proof of $\int gd\mu=\sum_{n=1}^\infty \mu (E_n)$ we can use MCT (Monotone Convergence Theorem).

Let $g_k=\sum_{n=1}^k I_{E_n}$ . Then $g_k\uparrow g$ . By MCT, $\int gd\mu=\lim_{k\to\infty} \int g_k d\mu=\lim_{k\to\infty} \sum_{n=1}^k \mu (E_n)=\sum_{n=1}^\infty \mu(E_n)$ .

Measurability of product fg

In the previous chapters, Bartle showed that that if f is in M(X,X), then the functions $cf, f^2, |f|, f^+, f^-$ are also in M(X,X).

The case of the measurability of the product fg when f, g belong to M(X,X) is a little bit more tricky. If $n\in\mathbb{N}$ , let $f_n$ be the “truncation of f” defined by $f_n (x)=\begin{cases}f(x), &\text{if }|f(x)|\leq n, \\ n, &\text{if } f(x)>n,\\ -n, &\text{if }f(x)<-n\end{cases}$

Let $g_m$ be defined similarly. We will work out the proof that $f_n$ and $g_m$ are measurable (Bartle left it as Exercise 2.K).

Proof:

Each $f_n$ is a function on $X$ to $\mathbb{R}$ .

$\{x\in X:f_n (x) >\alpha\}=\begin{cases}\{x \in X: f(x)>\alpha\}, &\text{if }-n<\alpha <n,\\ \emptyset, &\text{if }\alpha\geq n,\\X, &\text{if }\alpha\leq -n \end{cases}$

All of the above sets are in X.

Thus, we may use an earlier Lemma 2.6 to show that the product $f_n g_m$ is measurable.

We also have $f(x)g_m (x)=\lim_n f_n (x)g_m (x)$ , and using an earlier corollary that says that if a sequence $(f_n)$ is in M(X,X) converges to f on X, then f is also in M(X,X), we have that $f(x)g_m (x)$ belongs to M(X,X).

Finally, (fg)(x)=f(x)g(x)= $\lim_m f(x)g_m (x)$ , and hence fg also belongs to M(X,X).

This is a very powerful result of Lebesgue integration, since we can see that the theory includes extended real-valued functions, and prepares us to integrate functions that can reach infinite values!

Source: The Elements of Integration and Lebesgue Measure

Measure and Integration Recommended Book

I have added a new addition to the Recommended Books for Undergraduate Math, which is one of my most popular posts!

The new book is The Elements of Integration and Lebesgue Measure, an advanced text on the theory of integration. At the high school level, students are exposed to integration, but merely the rules of integration. At university, students learn the Riemann theory of integration (Riemann sums), which is a good theory, but not the best. There are some functions which we would like to integrate, but do not fit nicely into the theory of Riemann Integration.

I am personally reading this book as well, as I didn’t manage to study it in university, but it is a key component for graduate level analysis. Students interested in advanced Probability (see this post on Coursera Probability course) would be needing Lebesgue theory too!