# A little confusion from the Lebesgue integral definition

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 $$\int f(x)dx=\infty$$

• $M$ and $N$ depend on $g$ and $g$ is not fixed. – conditionalMethod Oct 14 '19 at 11:11
• Consider the case where $f=1$ (constant) and the functions $g_k$=$\chi_{[-k,k]}$ – MPW Oct 14 '19 at 11:51

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 $$\int_\mathbb Rf(x)\,\mathrm dx=\infty$$.
• 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 $+\infty$ – Bubblethan Oct 14 '19 at 11:23