Is there a function with infinite integral on every interval?

Question

Could give some examples of nonnegative measurable function $f:\mathbb{R}\to[0,\infty)$, such that its integral over any bounded interval is infinite?

Answer 1

The easiest example I know is constructed as follows. Let $q_{n}$ be an enumeration of the rational numbers in $[0,1]$. Consider $$g(x) = \sum_{n=1}^{\infty} 2^{-n} \frac{1}{|x-q_{n}|^{1/2}}.$$ Since each function $\dfrac{1}{|x-q_{n}|^{1/2}}$ is integrable on $[0,1]$, so is $g(x)$ [verify this!]. Therefore $g(x) < \infty$ almost everywhere, so we can simply set $g(x) = 0$ in the points where the sum is infinite.

On the other hand, $f = g^{2}$ has infinite integral over each interval in $[0,1]$. Indeed, if $0 \leq a \lt b \leq 1$ then $(a,b)$ contains a number $q_{n}$, so $$\int_{a}^{b} f(x)\,dx \geq \int_{a}^{b} \frac{1}{|x-q_{n}|}\,dx = \infty.$$ Now in order to get the function $f$ defined at every point of $\mathbb{R}$, simply define $f(n + x) = f(x)$ for $0 \leq x \lt 1$.

Answer 2

We don't really need that $|x-q_n|^{1/2}$ construction first.

Let $q_n$ be an enumeration of all rational numbers. Define function $g: \mathbb{R} \rightarrow [0,\infty]$ by:

$$ g(x)= \begin{cases} \sum_{n=0}^\infty \frac{1}{4^n |x-q_n|}, & \text{if}\ x\notin\mathbb{Q} \\ 0, & \text{if}\ x\in\mathbb{Q} \\ \end{cases} $$

Define set $ E=\bigcup_{k=1}^\infty \bigcap_{n=1}^\infty \left\{ x\in \mathbb{R}:|x-q_n|\ge\frac{1}{2^n k} \right\} $. Then we have $\mu(\mathbb{R} \setminus E)=0$, and $g$ converges on $E$. Then, $f=\mathcal{X}_E \cdot g$ is finite on its domain. Further, $\mu \text{-} a.e. f=g$, and thus $f$ has the same integral as $g$ over any non-empty open interval, which is infinite.

proof: For any non-empty open interval $I=(a,b)$, $\exists q_k \in (a,b) \cap \mathbb{Q}$, $$ \int_I f \textit{d} \mu = \int_{I} g \textit{d} \mu \ge \int_{I\setminus\mathbb{Q}} \textstyle \frac{1}{4^k |x-q_k|} \displaystyle \textit{d} \mu = \int_{I\setminus\{q_k\}} \textstyle \frac{1}{4^k |x-q_k|} \displaystyle \textit{d} \mu = \infty. $$

Answer 3

Let $I_1,I_2,I_3,\cdots$ enumerate the open intervals with rational endpoints. Construct pairwise disjoint closed nowhere dense sets of positive measure (i.e. fat Cantor sets) $A_n\subset I_n$; this can be done since $A_1\cup\cdots\cup A_{n-1}$ is nowhere dense. So every interval contains infinitely many $A_n$'s. Define $f:\mathbb R\to\mathbb R$ so that $f(x)=1/\lambda(A_n)$ if $x\in A_n$ and $f(x)=0$ if $x\notin\bigcup_nA_n$.