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I want to prove the following result:

If $\left( X, \mathcal{B}, \mu \right)$ is a $\sigma$-finite measure space and $f: X \rightarrow \mathbb{R}$ is a measurable function, then there is a sequence of integrable simple functions which converges pointwise to $f$.

To do so, I am given the hint as contruct $\phi_n: X \rightarrow \mathbb{R}$, as

$$\phi_n = \chi_{X_n} \left( \sum\limits_{k = 1}^{n2^n} \dfrac{k}{2^n} \chi_{f^{-1} \left[ \frac{k}{2^n}, \frac{k + 1}{2^n} \right]} + n \chi_{f^{-1} \left[ n, \infty \right)} \right).$$

Here, $X = \bigcup\limits_{n = 1}^{\infty} X_n$, where $\mu \left( X_n \right) < \infty$ for each $n \in \mathbb{N}$. Now, I can see that each $\phi_n$ is a simple function (since it can take only finitely many values and is defined on measurable sets. Also, it is integrable with the estimate

$$\int\limits_X \phi_n \mathrm{d}\mu = n \mu \left( X_n \right) \left( n 2^{n - 1} + 1 \right).$$

However, what I cannot see (directly) is that why should $\phi_n \to f$ pointwise?

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1 Answer 1

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There are some typos in your definition of $\phi_n$. First of all, this works only for non-negative measurable functions $f$ since your $\phi_n$'s are all non-negative. The sum starts from $k=0$. Also, you cannot include both the end points so I will drop the right end-point in this proof.

For any $x$ note that $f(x) <n$ for $n$ sufficienty large. For such $n$ there exists a unique $k$ such that $\frac k {2^{n}} \leq x< \frac {k+1} {2^{n}}$. This implies $k \leq n 2^{n}$. Now $\frac k {2^{n}} \leq x< \frac {k+1} {2^{n}}$ and $\phi_n(x)=\frac k {2^{n}}$. Thes two inequalities give $|f(x)-\phi_n(x)|$ is less than the length of the interval $[\frac k {2^{n}}, \frac {k+1} {2^{n}})$ which is $\frac 1 {2^{n}}$. Hence $\phi_n(x) \to f(x)$. If $f$ is also integrable then so is each $\phi_n$ since $0 \leq \phi_n \leq f$.

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  • $\begingroup$ For an arbitrary real measurable function (not necessarily non-negative), what can be done? Can we consider $f = f^+ - f^-$, where $f^+ = \max \left\lbrace f, 0 \right\rbrace$ and $f^- = \max \left\lbrace -f, 0 \right\rbrace$ and apply this result? We will then get two sequences, say $\left( \phi_n \right)_{n \in \mathbb{N}}$ and $\left( \psi_n \right)_{n \in \mathbb{N}}$. Then, we can construct our final sequence as $\alpha_n = \phi_n - \psi_n$. Is this approach correct? $\endgroup$ Commented Dec 15, 2020 at 11:47
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    $\begingroup$ All you have to do is change $f^{-1}([n,\infty))$ to $f^{-1}([n,\infty)\cup (-\infty,-n])$ and the sum over $k$ from $-\infty$ to $+\infty$. @AniruddhaDeshmukh $\endgroup$ Commented Dec 15, 2020 at 11:49
  • $\begingroup$ Also, why have we considered $\chi_{X_n}$? How is this playing a role in pointwise convergence? In your proof, how do we know that once we fix $x \in X$, we end up in the required $X_n$ (as per our choice of $n$)? $\endgroup$ Commented Dec 15, 2020 at 11:51
  • $\begingroup$ @AniruddhaDeshmukh The intervals $[\frac k {2^{n}}, \frac {k+1} {2^{n}})$ form a partition of the line. Any real number belongs to exactly one of these intervals (for fixed $n$). $\endgroup$ Commented Dec 15, 2020 at 11:54
  • $\begingroup$ Yes, I understand that. My question is, what would happen if $x \in X_{n_0}$ and the choice where $f \left( x \right) < n$ is such that $n \neq n_0$? How can we still use it? $\endgroup$ Commented Dec 15, 2020 at 11:57

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