3
$\begingroup$

Let $(X,\mathcal F,\mu)$ be measurable space with $\mu(X)<\infty$. $\mathcal F$ is $\sigma$-algebra on X and $\mathcal F$ is generated by algebra $\mathcal F_0$. Prove that for every measurable function $f$ defined on $X$ and $\epsilon>0$, there exists a simple function $f_{\epsilon}=\sum_{k=1}^nc_k 1_{A_k}$ with $A_k\in \mathcal F_0$ such that $\mu\{x: |f(x)-f_{\epsilon}(x)|\geq \epsilon\}<\epsilon$, $c_k$ is constant.

I have no idea to proceed the proof. Please give me some hints. Thanks

$\endgroup$
  • $\begingroup$ Recall that a.e. convergence implies convergence in measure in finite measure spaces. I would start with an increasing sequence of simple functions (not restricted to $\mathcal F_0$) which converges to $f$ pointwise, then approximate one of these functions with a sequence of simple functions restricted to $\mathcal F_0$. $\endgroup$ – Math1000 Aug 20 '15 at 19:33
1
$\begingroup$

Hints:

  1. Since $\mathcal{F}_0$ is an algebra, we have $X \in \mathcal{F}_0$. Define $$\mathcal{D} := \{F \in \mathcal{F}; 1_F \, \text{satisfies the claim}\}.$$
  2. Show that $X \in \mathcal{D}$ and $$F \in \mathcal{D} \implies X \backslash F \in \mathcal{D}.$$
  3. Let $(F_j)_{j \in \mathbb{N}} \subseteq \mathcal{D}$ be pairwise disjoint and fix $\epsilon>0$. Set $F := \bigcup_{j \geq 1} F_j$ and $f(x) := 1_F(x)$. Since $F_j \in \mathcal{D}$ there exists $$f_{\epsilon}^j := \sum_{k=1}^{n_j} c_k^j 1_{A_k^j}$$ such that $\mu(|1_{F_j}-f_{\epsilon}^j| \geq \epsilon 2^{-(j+1)}) <\epsilon 2^{-(j+1)}$. Choose $N \in \mathbb{N}$ sufficiently large such that $$\mu \left( F \backslash \bigcup_{j=1}^N F_j \right) \leq \frac{\epsilon}{2}$$ and set $$f_{\epsilon} := \sum_{j =1}^N f_{\epsilon}^j.$$ Conclude from $$\begin{align*}\mu(|f-f_{\epsilon}| \geq \epsilon) &\leq \mu \left( \left| 1_F - 1_{\bigcup_{j=1}^N F_j} \right| \geq \frac{\epsilon}{2} \right) + \mu \left( \left|1_{\bigcup_{j=1}^N F_j}-f_{\epsilon}\right| \geq \frac{\epsilon}{2} \right) \\ &\leq \mu \left( F \backslash \bigcup_{j=1}^N F_j \right)+ \sum_{j =1}^N \mu(|1_{F_j}-f_{\epsilon}^j| \geq \epsilon 2^{-(j+1)})\end{align*}$$ that $F \in \mathcal{D}$.
  4. Step 2&3 show that $\mathcal{D}$ is a Dynkin system. As $\mathcal{F}_0 \subseteq \mathcal{D}$ is $\cap$-stable and $\sigma(\mathcal{F}_0)=\mathcal{F}$, we get $$\mathcal{F}=\sigma(\mathcal{F}_0) = \delta(\mathcal{F}_0) \subseteq \delta(\mathcal{D}) = \mathcal{D}.$$
  5. Now let $f \geq 0$ be a measurable function and fix $\epsilon>0$. Define $$B_j := \left\{ j \frac{\epsilon}{4} \leq f < (j+1) \frac{\epsilon}{4} \right\} \in \mathcal{F}, \qquad j \geq 0,$$ and set $$g(x) := \sum_{j \geq 0} j \frac{\epsilon}{4} \cdot 1_{B_j}(x).$$ By construction, $$\mu(|f-g| \geq \epsilon/2)=0.$$ Using step 4 and a very similar argumentation as in step 3, we find a simple function $$f_{\epsilon} = \sum_{k=1}^n c_k 1_{A_k}$$ such that $$\mu(|g - f_{\epsilon}| \geq \epsilon/2) < \frac{\epsilon}{2}.$$ Conclude from the triangle inequality that $\mu(|f-f_{\epsilon}| \geq \epsilon)<\epsilon$.
  6. For arbitrary $f$ measurable write $f= f^+-f^-$ where $f^+$ and $f^-$ denote the positive part and negative part of $f$, respectively.
$\endgroup$
  • $\begingroup$ I know that this is an old question, but since I'm struggling with a smiliar question, I need to ask: In your third step, I don't see why $f_\varepsilon$ should be a simple function. So, how do you conclude $F\in\mathcal D$? (And let me advertise my question about the similar problem) $\endgroup$ – 0xbadf00d Jan 15 '18 at 21:59
  • $\begingroup$ @0xbadf00d You are right; there was something off about the third step. Please see the edited answer... $\endgroup$ – saz Jan 16 '18 at 18:03
  • $\begingroup$ I think the crucial point is the measure approximation $\mu \left( F \backslash \bigcup_{j=1}^N F_j \right) \leq \frac{\epsilon}{2}$. Did you take a look at my other question? There I'm basically asking for the same result, but for approximation in the sense of pointwise convergence. $\endgroup$ – 0xbadf00d Jan 16 '18 at 22:10
  • $\begingroup$ @0xbadf00d The above result implies in particular that there exists a sequenc of simple functions $f_n$ such that $f_n \to f$ $\mu$-almost everywhere. I don't think that it is possible to discuss away the exceptional null set (... I don't have a conterexample, it's just a feeling). $\endgroup$ – saz Jan 17 '18 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.