Let $A := \{(x, y, z) ∈ R^3 : 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ 1\}$ and $f : R^3 → R$ a function defined by $f(x, y, z) :=\begin {cases} \frac {sinx} x ,\; for\; x \ne 0,\\ 1 , \; for\; x = 0\end {cases}$ . Show that f is measurable and integrable on A
f is measurable if
- It's domain is measurable
- for any a belonging to R one of the following is satisfy
x|f(x)>a is measurable
$x|f(x)\geq a$ is measurable
x|f(x)<a is measurable
$x|f(x)\leq a$ is measurable
- So the domain would be [0,1]? so it would be measurable
We let a be a real number, and B={x∈R:f(x)<a}. Write out the intervals $I_1=(0,1]$, $I_2=(0)$ $B_k=B\cap I_k$ for k∈{1,2}
B is the disjoint union of the $B_k$ subsets.
Then x is in $B_1\text {iff}\; x\in I_1\; and\; \frac{sinx} x<a,\; i.e.\; x\in(0,k\pi]\cap I_1$ Hence, $B_1=(o,k\pi]\cap I_1$ is a measurable set since it's the intersection of two intervals?
Now to $B_2 : x$ is in B2 iff x∈I2 and a=1 So B2=I2 if a=1 and B2=∅ if a≠1. It's a measurable set in both cases. B is measurable as a (finite) union of measurable sets.
Is this correct? I used an answer to a similar question as a template, but I am not sure if I applied it correctly.
Integrable: f is integrable on A $x\in (0,1]$ t f is (Riemann) integrable on [a, b] if the upper integral and lower integral of f on [a, b] are equal; that is, $(U)\int^b_af = (L)\int^b_a f$.
I don't know how to apply it because I get that the integral is a sequence $(x-\frac{x^3} {3(3!)}+\frac{x^5} {5(5!)}-...)$ Not sure how to move on from here.
Thank you in advance for the help!
Domain isn't $[0, 1]$, it's $A$ (by definition). That $f$ doesn't depend on other variables doesn't change its domain.
For proving measurability and integrability I think it's simpler to first prove that $f$ is just continuous (well known fact that $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$, and outside of $x = 0$ it's just fraction of two continuous functions both of which are non-zero), and then refer to general result for continuous functions.
Also, given that you are asked about measurability, integrability probably means Lebesgue integrability, not Riemann.