Every $f\in L_{1}$ can be approximated by simple functions.

Question

Let $(X,\mathcal{X},\mu)$ a measure space. I want to prove that, for every $f\in L_{1}$, $\epsilon>0$, exists a simple function $\phi$ such that $$||\phi-f||_{1}<\epsilon.$$

Then I need to extends this to $L_{p},$ and ask if that is true for $L_\infty.$

Well, I can see intuitively that this is true. Since $f\in L_{1}$, I can assume that $$\mu(\{x\in X:|f(x)|=\infty\})=0.$$

Coming to the definition of integral, $|f|$ is integrable, so

$$\int|f|d\mu=\sup \int\phi d\mu, $$ where the sup is taken over the simple functions $\phi$ satisfying $0\leq \phi\leq |f|.$

So, for all $\epsilon >0,$ exists $\phi$ simple such that $$\int|f|d\mu-\int\phi d\mu <\epsilon\implies \int |f|-|\phi|d\mu<\epsilon\implies \int |f-\phi|d\mu<\epsilon\implies||\phi-f||_{1}<\epsilon,$$

because $\phi=|\phi|,$ and $\phi\leq |f|\implies |f-\phi|=|f|-\phi.$

Is my proof correct?

If yes, how can I extend this to $L_{p}?$

And what can I say about $L_{\infty}?$

Answer 1

It is not true that $|f-\phi| =|f|-\phi$. Yoru argument works fine when $f$ is non-negative. Now write $f$ as $f=f^{+}-f^{-}$ and get simple functions $\phi_1,\phi_2$ with $\|f^{+} -\phi_1\|<\epsilon /2$, $\|f^{-} -\phi_2\|<\epsilon /2$. Take $\phi=\phi_1-\phi_2$.

For $L^{p}$ start with $f \geq 0$ and choose simple functions $\phi_n$ increasing to $f$. Apply DCT to show that $\|\phi_n -f\|_p \to 0$. [ Note that $|f-\phi_n|^{p} \leq |f|^{p} $ which is integrable].

If $f \in L^{\infty}$ and $f \geq 0$ then there exits an $M$ such that $f \leq M$ almost everywhere. Define $\phi_n(x)=\frac {i-1} {2^{n}}$ when $\frac {i-1} {2^{n}} \leq f(x) <\frac i {2^{n}}$. Note that $\frac {i-1} {2^{n}} \leq f(x) \leq M$ so $i \leq 1+2^{n}M$. Hence $\phi_n$ is in $L^{\infty}$. Since $|\phi_n(x)-f(x)| \leq \frac 1 {2^{n}}$ by definition we get $\|\phi_n-f\|_{\infty} \to 0$.

Answer 2

\begin{align*} \int|f|-\phi<\epsilon \end{align*} does not entail \begin{align*} \int|f-\phi|<\epsilon, \end{align*} but the reverse direction holds true.

Anyway, to obtain the construction, for $1\leq p<\infty$, one assumes that $f$ is bounded, for \begin{align*} \|f-f\chi_{|f|\leq n}\|_{L^{p}}\rightarrow\infty. \end{align*} For bounded function $f$, we assume that $f\geq 0$, for $f=f^{+}-f^{-}$ and one can approximate both $f^{+}, f^{-}$.

For $0\leq f\leq M$, consider the simple functions \begin{align*} \varphi_{n}=\sum_{i=1}^{2^{n}}\dfrac{M(i-1)}{2^{n}}\chi_{f^{-1}(M(i-1)2^{-n},Mi2^{-n}]}, \end{align*} and use Lebesgue Dominated Convergence Theorem to conclude.

The above construction also holds for $p=\infty$, as we can take $M=\|f\|_{L^{\infty}}$ and \begin{align*} 0\leq f(x)-\varphi(x)\leq\dfrac{M}{2^{n}},~~~~\mu\text{-a.e.} \end{align*}