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I am just trying to make sure I have the definitions straight for the Lebesgue integral in the various categories of functions -- simple, nonnegative, etc.

I know that if $\varphi(x)=\sum_{k=1}^M c_k\chi_{F_k}$ is a simple function (where the $F_k$ are measurable and disjoint), then we simply define $\int_{\mathbb{R}^d}\varphi(x)dx=\sum_{k=1}^M c_km(F_k)$.

Then, for a function $f$ bounded by $M$ and supported by a set of finite measure E, we define $\int f(x)dx=\lim_{n\rightarrow\infty}\int\varphi_n(x)dx$, where $\{\varphi_n\}$ is a sequence of simple functions bounded by $M$ and supported by $E$.

Then, for nonnegative functions $f$, we define $\int f(x)dx=\sup_g\int g(x)dx$ where the supremum is taken over all measurable functions $g$ such that $0\leq g\leq f$, where $g$ is bounded and supported on a set of finite measure.

Now, with these definitions in mind, suppose I want to determine the integral $\int f(x)dx$ where $f(x)=n$ for some positive constant $n$. How would I actually go about doing this? It seems like I would appeal to the third situation above, but it doesn't seem clear to me how to construct an appropriate sequence.

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    $\begingroup$ Is $f(x)=n$ for all $x\in\Bbb R$? $\endgroup$ – Maximilian Janisch Jul 11 at 17:47
  • $\begingroup$ yes, that is what I was thinking. $\endgroup$ – ponchan Jul 11 at 17:49
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(Assuming that $f(x)=n>0$ for all real numbers $x$.)

For $m\in\Bbb N$, let $g_m := n \cdot \chi_{[-m,m]}$. Then all $g_m$ are bounded, simple, supported on a set with finite measure, and $0\le g_m\le f$.

So, by definition of the supremum, $$\int_{\Bbb R} f \geq \int_{\Bbb R} g_m = n \cdot 2m$$ for all natural numbers $m$. Since the right-hand side gets arbitrarily large as $m\to\infty$, we have $\int_{\Bbb R} f=\infty$ or undefined.

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