Consider the function $$f:[0,\infty)\to\mathbb{R},\qquad x\mapsto e^{-x}.$$ In freshmen calculus, we say that it is "integrable" because the improper integral $$\int_0^\infty e^{-x}\,dx$$ converges. This is basically saying that the Riemann integral converges. My question is: Is $f$ Lebesgue-integrable? To use the definition of Lebesgue-integrability, I need to decide whether there is an increasing sequence of nonnegative simple functions converging to $f$ pointwise and whose integrals don't blow up, but this seems like a too open-ended problem...
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Yes it is integrable. As in the integral is defined, and has a finite value. You do not need to explicitly find a sequence of simple functions to show this.
Firstly, the integrand is continuous hence measurable, and non-negative hence the integral makes sense but is possibly infinite. To prove it is finite, we can use the upper bound $$ x\in [n,n+1)\implies e^{-x} \le e^{-n}, \text{ i.e. }e^{-x} \le \sum_{n=0}^\infty \chi_{[n,n+1)}(x)e^{-n} $$ so that from $f\le g\implies \int f\le \int g$ and the convergence of geometric series, $$ \int_{[0,\infty)} e^{-x} dx \le \sum_{n=0}^\infty e^{-n} < \infty.$$
Calvin Khor
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