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Lebesgue integral of non-negative functions is defined as:

$$\int f(x)dx=\sup_g\int g(x)dx$$

where the sup is taken over all measurable functions $g$ such that $0\leq g\leq f$ and $g$ bounded, $m(supp(g))<\infty$.

My Questions: Suppose $g(x)\le M, m(supp(g))\le N$, then $\int g(x)dx\le M\cdot N$ should hold. But why would it be possible that $\int f(x)dx=\infty$

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    $\begingroup$ $M$ and $N$ depend on $g$ and $g$ is not fixed. $\endgroup$ – conditionalMethod Oct 14 '19 at 11:11
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    $\begingroup$ Consider the case where $f=1$ (constant) and the functions $g_k$=$\chi_{[-k,k]}$ $\endgroup$ – MPW Oct 14 '19 at 11:51
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For each function $g$ such that $0\leqslant g\leqslant f$ and that $g$ is bounded, we do have$$\int g(x)\,\mathrm dx\leqslant m\bigl(\operatorname{supp}(g)\bigr)\times\sup(g)$$indeed. But $\int f(x)\,\mathrm dx$ is the supremum of all these numbers and so, even eif each such number is a non-negative real number, that supremum may well be $\infty$. Simply take $f\colon\mathbb R\longrightarrow\mathbb R$ defined by $f(x)=1$ and, for each bounded measurable set $A\subset\mathbb R$, take $g=\chi_A$. Then $0\leqslant g\leqslant f$ and $\int_\mathbb Rg(x)\,\mathrm dx=m(A)$. So, yes, each $\int g(x)\,\mathrm dx$ is a non-negative real number, but $\int_\mathbb Rf(x)\,\mathrm dx=\infty$.

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  • $\begingroup$ Can I say it's similar to the natural numbers $\mathbb N$, every element in $\mathbb N$ is a specific number, but the supremum of natural numbers is $+\infty$ $\endgroup$ – Bubblethan Oct 14 '19 at 11:23
  • $\begingroup$ Yes. It is a similar situation. $\endgroup$ – José Carlos Santos Oct 14 '19 at 11:27

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