Integrable function $f$ and simple function $\phi$ such that $ \int{|f-\phi|} \> d\mu < \epsilon.$

Question

I am seeking solution verification for the following problem.

Suppose $f$ is an integrable function. We wish to show that there exists a simple function $\phi$ such that $$ \int{|f-\phi|} \> d\mu < \epsilon.$$ Attempt: Write $f = f^{+} - f^{-}$. The functions $f^{+}$ and $f^{-}$ are non-negative measurable functions, and our assumption implies that there integrals are finite. These functions can be approximated by monotonically increasing sequences of simple functions. In particular, there exists a sequence of simple functions $(\psi_{n})$ that monotonically increases to $f^{+}$ and there exists a sequence of simple functions $(\psi^{'}_{n})$ that monotonically increases to $f^{-}$. The Monotone Convergence Theorem implies $$\lim \int{\psi_{n}} \> d\mu = \int{f^{+}} \> d\mu$$ and $$\lim \int{\psi^{'}_{n}} \> d\mu = \int{f^{-}} \> d\mu.$$

Now, take $N$ large enough so that $n \geq N$ implies $$\left| \int{f^{+}} \> d\mu - \int{\psi_{n}} \> d\mu \right| < \frac{\epsilon}{2}.$$ Likewise, take $N'$ large enough so that $n \geq N'$ implies $$\left|\int{f^{-}} \> d\mu - \int{\psi^{'}_{n}} \> d\mu \right| < \frac{\epsilon}{2}$$ Observe that $|\int{f^{+}} \> d\mu - \int{\psi_{n}} \> d\mu| = \int{f^{+}} \> d\mu - \int{\psi_{n}} \> d\mu = \int{f^{+}-\psi_{n}} \> d\mu$ because $\psi_{n} \leq f^{+} $ for each $n$ and the linearity of the integral. The analogue is true for $f^{-}$ and $\psi^{'}_{n}$.

Put $\phi = \psi_{N} - \psi^{'}_{N'}$. Then,

\begin{align} \int{|f-\phi|} \> d\mu &= \int{|f^{+}-f^{-}-\psi_{N} + \psi^{'}_{N'}|} \> d\mu \\ &\leq \int{|f^{+}-\psi_{N}|+|-f^{-} + \psi^{'}_{N'}|} \> d\mu \\ &= \int{|f^{+}-\psi_{N}|+|-1||f^{-} - \psi^{'}_{N'}|} \> d\mu \\ &=\int{|f^{+}-\psi_{N}|+|f^{-} - \psi^{'}_{N'}|} \> d\mu \\ &= \int{f^{+}-\psi_{N}} \> d\mu +\int{f^{-} - \psi^{'}_{N'}} \> d\mu \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align} The proof is complete.

Answer 1

Actually, you are a little bit mistaken. Your proof fails when you claim that $\int|f^+ - \psi_N | d\mu < \frac{\epsilon}{2} $. What you get is $ | \int f^+ - \psi_N d\mu | < \frac{\epsilon}{2} $. To solve this problem you have to use the Dominated Corvenge Theorem as follows:

First let $f$ be a nonnegative function. Then there exists a monotone increasing sequence of simple functions $(\phi_n)$ such that $\phi_n \rightarrow f$. Hence $|\phi_n| \leq |f|$. Not only that, but $\phi_n - f \rightarrow 0 $. Now, notice that: \begin{equation} |\phi_n - f| \leq |\phi_n| + |f| \leq 2|f|. \end{equation}

But one has that $2|f|$ is integrable. Then, $|\phi_n - f|$ is integrable. But notice that $|\phi_n - f| \rightarrow 0 $. Being $|\phi_n - f|$ dominated by an integrable function and converging to $0$, you can apply the Dominated Convergence Theorem to see that:

\begin{equation}0 = \int 0 d\mu = lim \int |\phi_n - f| d\mu. \end{equation}

Now you can see that, if $f$ is a nonnegative function, there exists a simple function that is a good approximation for $f$.

The exercise follows observing that any function $f$ can be written as the difference between its postive and its negative parts.

Answer 2

One can just use the definition of the Lebesgue integral without having to rely on the dominated convergence theorem.

Indeed, there exists a simple function $s_1$ such that $0 \leq s_1 \leq f^{+}$ and $\int f^+ d\mu < \int s_1 d\mu+\frac{\epsilon}{2}$. Similarly there exists a simple function $s_2$ such that $0\leq s_2\leq f^-$ and $\int f^- d\mu < \int s_2 d\mu + \frac{\epsilon}{2}$. Since $f^+-s_1 \geq 0$ and $f^--s_2\geq0$ we get $\int |f-s_1-s_2| d\mu \leq \int |f^+-s_1| d\mu +\int |f^--s_2|d\mu = \int f^+ -s_1 d\mu + \int f^--s_2 d\mu < \epsilon$.