I'm asked to find the Lebesgue integral of sin (1/x) on the interval (0,1], if it exists. But is sin(1/x) even a measurable function? It seems to me that the answer is no, but I cannot prove this.
It is continuous on $\displaystyle (0,1]$, hence it is of course measurable. In addition, it is bounded, so that it is Lebesgue-integrable on this bounded interval (Indeed, $\displaystyle \left|\sin\frac{1}{x}\right|\leq 1$ and $\displaystyle x\mapsto 1$ is integrable on $\displaystyle (0,1]$, so that $\displaystyle x\mapsto\sin\frac{1}{x}$ is integrable on $\displaystyle (0,1]$ as well).
Integration by parts gives:
$\displaystyle \int \sin(1/x)\ dx = x \sin\left(\frac{1}{x}\right)-\text{Ci}\left(\frac{1}{x}\right) + C$
So:
$\displaystyle \int_0^1 \sin(1/x)\ dx = \sin(1)-\text{Ci}(1)-\lim_{a \to 0}\left[ a\sin\left(\frac{1}{a}\right)-\text{Ci}\left(\frac{1}{a}\right)\right]$ $\displaystyle =\sin(1)-\text{Ci}(1)\approx 0.504067 $
CB
Ok thanks very much for this. I see now that I was being a bit silly when I suggested that sin(1/x) was probably not measurable.
Perhaps this is a better question to highlight the problems I'm having with Lebesgue integration:
Is f(x) tan x integrable over the interval (0,pi/2)?
Now we can show that f is measurable on the interval, and hence continuous. However, it is unbounded and hence not necessarily integrable. I know we can say " $\displaystyle \int tan x = \log \sec x$ " and probably this to show the integral doesn't exist, but this isn't really the point. I'm after something that show from first principals or standard results (eg Monotone convergence theorem etc) why this isn't integrable over the stated interval. Thanks.