Measurability of a function in $\mathcal{B}(\mathcal{C}([0,1],\mathbb{R}))$

Question

The Question I cant answer is, why is $\Lambda_a:\mathcal{C}([0,1],\mathbb{R})\rightarrow\mathbb{R}$, given by $\Lambda_a(\omega):=\lambda(\{t \in [0,1]:\omega(t)>a\})$ is $\mathcal{B}(\mathcal{C}([0,1],\mathbb{R}))$ measurable.

I can see that $\{t \in [0,1]:\omega(t)>a\}$ is an Open Set for all continuous functions $\omega$, but that's about it.

I don't even know what I can do with the Hint, to look at the function $h:\mathcal{C}([0,1],\mathbb{R})\times [0,1] \rightarrow \mathbb{R}$ defined by $h(\omega,t)=\omega(t)$.

If I look at $\{h>a\}$ it would yield all combinations of $\omega$ and $t$ with $\omega(t)>a$ and if $h$ is measurable in $\mathcal{B}(\mathcal{C}([0,1],\mathbb{R}))\oplus\mathcal{B}([0,1])$ (I Hope this is the right construction) I would get open Sets in $\mathcal{C}([0,1],\mathbb{R})$ and $[0,1]$. But how would I proceed if this is even correct?

thank you for your time

Edit: $\lambda$ is the Lebesgue measure

Answer 1

For simplicity of notation, let $X\equiv[0,1]$ and $C\equiv C([0,1],\mathbb R)$.

Claim: The function $h:C\times X\to\mathbb R$ defined as $h(\omega,t)=\omega(t)$ is continuous.

Proof: Fix $\omega\in C$, $t\in X$, and $\varepsilon>0$. Let $\delta>0$ be such that $|t'-t|<\delta$ implies that $|\omega(t')-\omega(t)|<\varepsilon/2$ (the existence of such $\delta$ is implied by the continuity of $\omega$). Then, if $\|\omega'-\omega\|<\varepsilon/2$ and $|t'-t|<\delta$, then $$|h(\omega',t')-h(\omega,t)|=|\omega'(t)'-\omega(t)|=|\omega'(t')-\omega(t')|+|\omega(t')-\omega(t)|\leq\|\omega'-\omega\|+\frac{\varepsilon}2<\varepsilon.\,\blacksquare$$

Let $H_a\equiv h^{-1}((a,\infty))$. Since $h$ is continuous, $H_a$ is an open (and hence Borel, considering the product $\sigma$-algebra; see the remark at the bottom) set in $C\times X$. For each $\omega\in C$, define $H_a^{\omega}$ to be the following set: $$H_a^{\omega}\equiv\{t\in X\,|\,(\omega,t)\in H_a\}.$$ This set is measurable (Proposition 2.34 in Folland, 1999). Moreover, the map $K:C\to \mathbb R$, defined as $$K(\omega)\equiv\lambda(H_a^{\omega})$$ is measurable (Theorem 2.36, ibid).

(A little care must be taken in using the last theorem, since it is applicable only if both measure spaces $(C,\mathscr{B}_C,\mu_{C})$ and $(X,\mathscr B_{X},\lambda)$ are $\sigma$-finite. This is clearly true for the latter, and we never use the properties of the measure $\mu_C$ on $(C,\mathscr{B}_C)$, so we might safely assume that $\mu_C$ vanishes identically.)

Now, let $U$ be a measurable set in $\mathbb R$. We wish to prove that $$\Lambda_a^{-1}(U)\equiv\left\{\omega\in C\,\Big|\,\lambda\left(\{t\in X\,|\,\omega(t)>a\}\right)\in U\right\}$$ is measurable. But this set is equal to: \begin{align*} &\left\{\omega\in C\,\Big|\,\lambda\left(\{t\in X\,|\,h(\omega,t)>a\}\right)\in U\right\}=\left\{\omega\in C\,\Big|\,\lambda\left(\{t\in X\,|\,(\omega,t)\in H_a\}\right)\in U\right\}\\=&\left\{\omega\in C\,\Big|\,\lambda\left(H_a^{\omega}\right)\in U\right\}=\left\{\omega\in C\,\Big|\,K(\omega)\in U\right\}=K^{-1}(U). \end{align*} Since $K$ is a measurable map and $U$ is a measurable set, $K^{-1}(U)$ is measurable.

ADDED: You might say that I wasn't careful enough to distinguish $\mathscr B_C\otimes \mathscr B_X$ from $\mathscr B_{C\times X}$. In general, this may be a problem, because $\mathscr B_C\otimes \mathscr B_X\subseteq\mathscr B_{C\times X}$ and the inclusion may be strict. But this is not a problem here, since $X$ is separable and $C$ is also separable (consider the set of all polynomials on $[0,1]$ with rational coefficients and apply Weierstrass's uniform approximation theorem), which implies that $\mathscr B_C\otimes \mathscr B_X=\mathscr B_{C\times X}$.

Answer 2

Let's show that $\Lambda_a$ is lower semicontinuous. Since semicontinuity implies measurability the result will follow.

---

Let $f_0:[0,1]\to\mathbb{R}$ be continuous and consider $\epsilon>0$. Let us show that there is an open ball $B_{\delta}=B(f_0,\delta)$ centered at $f_0$ such that for all $f\in B_{\delta}$, $\Lambda_a(f)\geq\Lambda_a(f_0)-\epsilon$. There is nothing to prove if $\Lambda_a(f_0)\leq\epsilon$. So let us assume $\Lambda_a(f_0)>\epsilon$ Consider the decomposition $$X=\bigcup{}^{\uparrow}_{n\geq 1}X_n$$ where $X=\lbrace x\in[0,1]:f_0(x)>a\rbrace$ and for all $n\geq 1$, $X_n=\left\lbrace x\in[0,1]:f_0(x)>a+\frac1n\right\rbrace$. Then $X_n$ is an increasing sequence of open sets whose union is $X$. Since $\lambda(X_n)=\Lambda_a(f_0)$, the standard properties of Lebesgue measure imply that there exists $N$ such that $\lambda(X_N)\geq\Lambda_a(f_0)-\epsilon$ Then by definition $$\forall f\in B_{\frac1N},\Lambda_a(f)\geq\Lambda_a(f_0)-\epsilon$$ so that $\Lambda_a$ is lower semicontinuous.