Show that $e^{f(x)}$ is Riemann Integrable

Question

Suppose that $f:\mathbb{R} \to \mathbb{R}$ is Riemann Integrable and $f = 0$ for $f \notin [a,b]$. Show that $e^{f(x)}*\chi_{[a,b]}$ is Riemann Integrable.

I think this means that:

$g(x) = e^{f(x)} = e^{f(x)} $for$ a \leq x \leq b$ and $0$ otherwise

I am trying to find two step functions for showing this is a Riemann Integrable, I am trying to use $e^{(f(x))^2}$ as my upper step function but with little success, does anyone have any suggestions? Thanks

I am confused (syntactically and logically) by that $f=0$ condition — Hagen von Eitzen, Feb 13 '14 at 18:30

score 1 · Answer 1 · answered Feb 13 '14 at 18:29

1

Hint: To find step functions for $e^{f(x)}$, you may assume suitable step functions for $f(x)$ given.

answered Feb 13 '14 at 18:29

Hagen von Eitzen

374,180

So I assume $\phi(x)$ and $\psi(x)$ are upper and lower step functions of $f(x)$ respectively. I am trying to show the supremum of $\sum|e^{f(x)} -e^{f(y)}||I_j| < \epsilon$ (as this is a method I have used before to prove $f^2$ is Riemann Integrable) Is this the next step? – yhu Feb 13 '14 at 19:05

score 0 · Answer 2 · answered Feb 15 '14 at 04:44

We can use the following the result:

If $f(x)$ is Riemann integrable on $[a, b]$ and range of $f(x)$ is $[A, B]$ and $g(x)$ is continuous on $[A, B]$ then $g(f(x))$ is Riemann integrable on $[a, b]$.

The result is false if we don't assume continuity of $g$. Clearly if we assume the continuity of $g$ then we can see that the points of discontinuity of $g(f(x))$ are same as those of $f(x)$. Since $f(x)$ is Riemann integrable the discontinuities of $f(x)$ form a set of measure zero and therefore the discontinuities of $g(f(x))$ form a set of measure zero. It follows that $g(f(x))$ is Riemann integrable on $[a, b]$.

Here in the current question we have $g(x) = e^{x}$ which is continuous on any finite interval so that $e^{f(x)}$ is Riemann integrable.

Show that $e^{f(x)}$ is Riemann Integrable

2 Answers2