Lebesgue Integral of $e^x$

Question

I want to show that that $f_{\alpha}: ]0,\infty[ \to \mathbb{R}$, where $x \to e^{-\alpha x}$ for any $\alpha \gt 0, \alpha \in \mathbb{R}$ is Lebesgue integrable, i.e. $$\int_{]0,\infty[}abs(f_{\alpha})d\lambda \lt\infty$$ Since I am not allowed to calculate the integral with Riemannian methods, I want to write out the integral expliclitly. For that I need a sequence of simple functions converging to $f_{\alpha}$, something I cannot find. Otherwise, I could write $e^{-\alpha x}$ as a composition of $e^x$ and $-\alpha x$ and then use an according theorem, but there I would still need to calculate $e^x$. So this should look like $$\int_{]0,\infty[}f_{\alpha}d\lambda=sup_{k\in\mathbb{N}}\int_{]0,\infty[}f_{\alpha,k}d\lambda=sup_{k\in\mathbb{N}}\left(\sum_{i=0}^N\beta_i\lambda(A_i)\right)_{\alpha, k}$$with $\lambda$ being the ordinary Borel-Lebesgue measure. I am now looking for $f_{\alpha, k}=\sum_{i=0}^N\beta_i 1_{A_i}$. Any help greatly apppreciated!

Answer 1

Take $f_{\alpha,k} =\sum\limits_{j=1}^{k} \frac 1 {j^{2}} I_{A_j}$ where $A_j=(\frac {2 \ln \, j} {\alpha},\frac {2 \ln \, (j+1)} {\alpha})$. Note that if $x \in A_j$ then $e^{-\alpha\, x}$ lies between $\frac 1 {(j+1)^{2}}$ and $\frac 1 {j^{2}}$.