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For an arbitrary interval $I$, how can we find a positive on $I$ integrable function? And how does one construct a measurable but not integrable function. If not all measurable functions are integrable, what is the best way to determine if a function is L-Integrable?

thx in advance

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Positive integrable function?? What about the constant $1$? For the other question, for example if the integral of the positive and the negative part are both $(\pm)\infty$. –  Berci Feb 3 '13 at 13:26

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Probably you consider the lebesgue measure on the intervall $I \subseteq \mathbb{R}$...? In this case:

  1. Take any positive (lebesgue) integrable function $f: \mathbb{R} \to \mathbb{R}$ and restrict it to $I$, then it's a positive integrable function on $I$. For example $$f(x) := \frac{1}{x^2+1} \qquad (x \in I)$$ does the job.
  2. Let $x_0 \in I$ such that $B(x_0,\varepsilon) \subseteq I$ for some $\varepsilon>0$. Define $$g(x) := \begin{cases} \frac{1}{x-x_0} & x \in B(x_0,\varepsilon) \backslash \{x_0\} \\ 0 & x \notin B(x_0,\varepsilon) \vee x=x_0 \end{cases}$$ $g$ is measurable, but not integrable.
  3. A measurable function $f$ is integrable iff there exists a positive integrable function $g$ such that $$\forall x \in I: |f(x)| \leq g(x)$$ This means that you should try to find an integrable upper bound of $f$. In particular it's important to know some integrable functions, for example
    • $x \mapsto \frac{1}{x^\alpha} \in L^1([a,\infty))$ for all $a>0$, $\alpha>1$ and $x \mapsto \frac{1}{x^\alpha} \in L^1([0,a])$ for all $a>0$, $\alpha<1$
    • $x \mapsto e^{-x} \in L^1(\mathbb{R})$ and $x \mapsto p(x) \cdot e^{-x^2} \in L^1(\mathbb{R})$ for an arbritary polynomial $p$
    • $\ldots$
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