Measure Theory-Borel sets-Lebesgue integral-Monotonce Convergence Theorem question

I am preparing for an exam in measure theory and probabilities and the question below is from a previous exam in this course. I have tried to answer it, though I miss certain key points in my solution. I would appreciate any help. Thank you.

For all $A\in\mathcal{B}([-1,1])$, define: $-A:=\{x\in\mathbb{R}:-x\in A\}$. Let $\lambda$ denote the Lebesgue measure. Define: $\mathcal{A}=\{A\in\mathcal{B}([-1,1]):\lambda(A)=\lambda(-A)\}$.

i.) Prove that $\mathcal{A}$ is a d-system.

ii.) Show that $\mathcal{A}=\mathcal{B}([-1,1])$.

iii.) Show that if $f:[-1,1]\rightarrow\mathbb{R}$ is a $\mathcal{B}([-1,1])$ simple function and it holds that $f(x)=-f(-x)\forall x\in[-1,1]$ then: $\int_{-1}^1 f(x)dx=0$.

iv.)Show the same as above only for $f$ now Lebesgue integrable.

Now my answers so far are the following:

For question i.):

Pick $E,F\in\mathcal{A}$ s.t.: $E\subset F$. First note that $E,F\in\mathcal{B}([-1,1])$, hence $F\setminus E\in\mathcal{B}([-1,1])$. Now: $\lambda(F\setminus E)=\lambda(F)-\lambda(E)=\lambda(-F)-\lambda(-E)=\lambda(-F\setminus -E)=\lambda(-(F\setminus E))$.

Now consider $E_n\in\mathcal{A}$ such that: $E_n\subset E_{n+1}$. First we have that $\cup E_n\in\mathcal{B}([-1,1])$. Now $\lambda(\cup E_n)=\sum \lambda(E_n)=\sum \lambda(-E_n)=\lambda(-\cup E_n)$. Therefore $\cup E_n\in\mathcal{B}([-1,1])$.

For question ii.):

Pick $A\in\mathcal{A}$. Then by definition $A\in \mathcal{B}([-1,1])$.

Pick $B\in\mathcal{B}([-1,1])$. Then $\lambda(B)=\int_{B\cap[-1,1]}d\lambda=\int_{-B\cap [-1,1]}d\lambda=\lambda(-B)$. Hence $B\in\mathcal{A}$ and therefore $\mathcal{A}=\mathcal{B}([-1,1])$.

For question iii.):

Let $f$ be a $\mathcal{B}([-1,1])$ measurable simple function with $f(x)=-f(-x)\forall x\in[-1,1]$.

Then: $f(x)=\sum_{i=1}^{n}a_i1_{A_i},A_i\in [-1,1],a_i\in [0,+\infty)$.

Now

$\int_{-1}^{1}f(x)=\int_{-1}^{1}\sum_{i=1}^{n}a_i1_{A_i}=\sum_{i=1}^{n}a_i\lambda(A_i\cap [-1,1])$. Now since $A_i\in\mathcal{B}([-1,1])$, we can use question ii.) to get that: $\lambda (A_i)=\lambda (-A_i)$, which in turn we can somehow use to derive what we need. Only as you can see I lack the details here.

For question iv.):

We know that $\exists$ sequence of simple functions $(f_n)$ such that: $f_{n+1}>f_{n}$, with: $\lim\sup f_n=f$. Using Monotone Convergence Theorem, we derive that: $\int_{[-1,1]}f(x)dx=\lim\sup\int_{[-1,1]}f_n(x)dx=0$ (by previous question).

• I think for i) your conclusion should be that $\cup E_n\in\mathcal A$ ;) Jan 14 '15 at 21:42
• @Math1000 yeah this is a typo, sorry Jan 14 '15 at 21:45

I don't follow your argument for ii). Where do you use $A$? Why are you writing $B\cap[-1,1]$ when it is assumed that $B\subset[-1,1]$? I don't see how you conclude that $\int_B\mathrm d\lambda = \int_{-B}\mathrm d\lambda$.

For iii), write $f=\sum_{i=1}^n a_i\chi_{A_i}$ where $a_i\in\mathbb R$ and $A_i\in\mathcal A$. Then $-f=\sum_{i=1}^n-a_i\chi_{A_i}$, so $$-f(-x)=\sum_{i=1}^n-a_i\chi_{A_i}(-x)=\sum_{i=1}^n-a_i\chi_{-A_i}(x).$$ By definition of the Lebesgue integral, $$\int_{[-1,1]}f\mathrm d\lambda = \sum_{i=1}^n a_i\lambda(A_i)=\sum_{i=1}^n a_i\lambda(-A_i).$$ As $f(x)=-f(-x)$ it follows that $$\sum_{i=1}^na_i\lambda(-A_i)=\sum_{i=1}^n-a_i\lambda(-A_i)=-\sum_{i=1}^n a_i\lambda(-A_i).$$ This implies that $\int_{[-1,1]}f\mathrm d\lambda = 0$ (as for a real number $y$, $y=-y$ if and only if $y=0$).

• yes you are right, my argument in ii.) is really messed up Jan 14 '15 at 22:43

For (ii), note that you're dealing with the measure of a set, not an integral with respect to a measure.

I think you need to use the definition of the Lebesgue measure here, I.e.,

$\lambda(A)=inf_{\mathcal{I}(A)}\sum_{I_{k}\in\mathcal{I}(A)}l(I_k)$,

where $\mathcal{I}(A)$ is a finite open cover of the set $A$, $I_k=(a_k,b_k)$ are the intervals in this cover, and $l$ is the ordinary length of an interval $(l(I_k)=b_k-a_k)$.

Observe that $l(I_k)=l((a_k,b_k))=b_k-a_k=-(a_k-b_k)=-a_k-(-b_k)=l(-b_k,-a_k)=l(-I_k)$. Applying this to each $I_k$ in the cover $\mathcal{I}(A)$ yields the result, as $I_k\in\mathcal{I}(A)\Leftrightarrow-I_k\in\mathcal{I}(-A)$.