Prove $F$ is in $L^1$

Question

This is an old qualifier exam question at my school

Let $f \in L^{1}([0,\infty))$ and for $x\geq 0$, define

$F(x) = \int_{(x,\infty)} f(t) e^{x-t} dm(t) $

Show that $F \in L^{1}([0,\infty))$

The first thing that throws me off is $dm(t)$ I think its same as $dt$.

Here is my work so far

$\int_{[0,\infty)} |f(t)| d\mu$ $= \int_{[0,\infty)}|\int_{(x,\infty)} f(t) e^{x-t} dm(t)|d\mu $

$\leq \int_{[0,\infty)}[\int_{(x,\infty)} |f(t) e^{x-t}| dm(t)]d\mu$

$\leq \int_{[0,\infty)}[\int_{(x,\infty)} |f(t) || e^{x-t}| dm(t)]d\mu$

I am trying to write $\int_{[x,\infty)} |f(t) || e^{x-t}| dm(t)$ $= \int_{[x,t)} |f(t) || e^{x-t}| dm(t) + \int_{[t,\infty)} |f(t) | dm(t)$

I dont want to apply holders as it will give me $f$ in $L^2$. I would appreciate a hint. May be I am going in totally wrong direction.

Answer 1

We have to prove finiteness of $$I:=\int_{[0,+\infty)}\int_{[0,+\infty)}|f(t)|e^{x-t}\mathbf 1\{t\geqslant x\}\mathrm dm(t)\mathrm dm(x).$$ Using Fubini-Tonnelli's theorem (that is, switching the integral when the integrand is a non-negative function), we infer that $$I=\int_{[0,+\infty)}e^{-t}|f(t)|\int_{[0,t)}e^x\mathrm dm(x)\mathrm dm(t).$$