Let $A= \lbrace \phi \in L^{1}(X) \: | \: \phi \: \mbox{is simple} \rbrace$. Prove $A$ is dense in $L^{1}(X)$. Basically my idea is to prove that for every $f \in L^{1}(X)$ and every $\epsilon>0$ there is a $\phi \in A$ such $||f-\phi||_{1}< \epsilon $ in other words there is $\phi \in A$ contained in every neighborhood of $f \in L^{1}(X)$.
But $||f-\phi||_{1}< \epsilon $ means that $\int |f-\phi| d \mu < \epsilon $. So how can I prove this las inequality for every $\epsilon>0$. Thanks!!
In this answer Simple functions are dense in L1
I dont know why $$ \int |f^+-\varphi^+|\,d\mu<\frac{\varepsilon}{2}\text{ and }\int |f^--\varphi^-|\,d\mu<\frac{\varepsilon}{2} $$for any $\varepsilon>0$.
Actually one approximates $f^{+}$ by a simple function $\phi$： \begin{align*} \int|f^{+}-\varphi|<\epsilon/2, \end{align*} and then approximates $f^{-}$ with another one $\psi$: \begin{align*} \int|f^{-}-\psi|<\epsilon/2, \end{align*} then $\varphi+\psi$ is a simple function that approximates $f=f^{+}-f^{-}$: \begin{align*} \int|f-(\varphi+\psi)|=\int|(f^{+}-\varphi)+(f^{-}-\psi)|\leq\int|f^{+}-\varphi|+\int|f^{-}-\psi|<\epsilon. \end{align*}
For nonnegative function $g$, we have by definition that \begin{align*} \int g=\sup\left\{\int\phi: 0\leq\phi\leq g\right\}, \end{align*} where $\phi$ is any simple function, so $\displaystyle g$ can be approximated by some $0\leq\phi\leq g$ by arbitrary small: \begin{align*} \int g-\int\phi<\epsilon, \end{align*} but then $|g-\phi|=g-\phi$ because $\phi\leq g$, so \begin{align*} \int|g-\phi|=\int g-\int\phi<\epsilon. \end{align*}