# Lebesgue integral of a constant function

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.

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

## 1 Answer

(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.