We get a continuous function $g:[a,b]\to\Bbb R$, and we defined: $$f(x)=\begin{cases} g(x) & \text{ if $x\in[a,b]$} \\0 & \text{ if $x\not\in [a,b]$} \end{cases}$$ Prove that $f$ is Lebesgue integrable and $\int_{\Bbb R}f=\int_a^b g$.
I think that $f$ is semicontinuous, however I'm having some troubles while proving so. I considered cases:
- if $g(a)=g(b)=0$,
- if $g(a)=0$ but $g(b)\not=0$,
- if $g(b)=0$ but $g(a)\not=0$,
- if $g(a)<0<g(b)$,
- if $g(b)<0<g(a)$,
- if $0<g(a)<g(b)$,
- if $0<g(b)<g(a)$,
- if $g(a)<g(b)<0$,
- if $g(b)<g(a)<0$,
I'm having troubles with the cases $g(a)<0<g(b)$. I keep thinking in these two particular examples: $$h_1(x)=\begin{cases} x & x\in[-1,1] \\ 0 & x\not\in[-1,1] \end{cases}, h_2(x)=\begin{cases} \tan x & x\in[\frac{-\pi}{2},\frac{\pi}{2}] \\ 0 & x\not\in[\frac{-\pi}{2},\frac{\pi}{2}]\end{cases}$$ I get confused to see whether they're semicontinuous, in $h_1$ we get $h_1(a)=-1<0<1=h_1(b)$, but in $h_2$ $h_2(a)=-\infty<0<\infty=h_2(b)$. What do I do? Are these functions really semicontinous?
If $g(a)$ and $g(b)$ have opposite sign, then $f$ is not globally semicontinuous. If $g(a) < 0 < g(b)$, then $f$ is
but not continuous in either point, hence $f$ isn't globally semicontinuous.
If $a < c < b$, then $f$ is however semicontinuous on the two intervals $(-\infty,c]$ and $[c,+\infty)$. That suffices to show that it is integrable.