# Function between two measurable functions is measurable

Let $f$ be a function on $\mathbb{R}^n$. Assume that for any $\epsilon>0$, there exist measurable functions $g, h \in L^1(\mathbb{R}^n)$ such that $g(x) \leq f(x) \leq h(x)$ for all $x \in \mathbb{R}^n$ and $$\int_{\mathbb{R}^n} (h(x)-g(x))\, dx < \epsilon$$ Prove that $f$ is measurable on $\mathbb{R}^n$ and $f \in L^1(\mathbb{R}^n)$.

My first thought was to show that $f(x)$ is a pointwise limit of measurable functions; but since all information we have about $f(x)$ is in $L^1$, it would be hard to consider pointwise limits.

My problems are:

(1) I am not sure how to prove $f$ is measurable. To establish measurability of $f(x)$, I need to go through the definition that $\{x: f(x)<c\}$ is measurable for all $c$. Now $\{x: f(x)<c\} = \{x: \exists h \, \text{measurable s.t.}\, f(x)\leq h(x)<c\}$, and the trouble is I am not sure how to express the latter as a countable union of measurable sets.

(2) I was wondering if the following proof for $f \in L^1$ is correct: Assuming $f$ is measurable, we want to show that $f \in L^1(\mathbb{R}^n)$. For any $n\in \mathbb{N}$, there is $h_n, g_n \in L^1$ with $h_n(x)\leq f(x) \leq g_n(x)$ for all $x$ and $\int |f(x)-h_n(x)|\, dx = \int f(x)-h_n(x)\, dx \leq \int g_n(x)-h_n(x)\, dx < 1/n$. Hence $||f-h_n||_{L^1} < 1/n$, and thus $f \in L^1$ by completeness of $L^1$.

In summary, I appreciate any help/hint on measurability of $f(x)$, and check on if the proof in (2) is correct. Thank you!

• It suffices to prove it for $g = 0$. – Gribouillis Aug 30 '17 at 18:23
• I think your proof for (2) is OK, but you don't really need completeness of $L^1$ if you already know that $f$ is measurable. You established that $\int |f - h_n| < 1/n$, and this implies that $\int |f| \leq \int |f - h_n| + \int |h_n| \leq 1/n + \int |h_n|$, which is finite since $h_n \in L^1$. – Bungo Aug 30 '17 at 18:35

There is a very useful result that you almost certainly proved on the way towards proving the completeness of $L^p(\mu)$:

If $u_n(x)$ is a sequence in $L^p(\mu)$ converging to $u(x)$ with respect to the $L^p(\mu)$ norm, then there exists a subsequence $u_{n_m}(x)$ of $u_n(x)$ that converges to $u(x)$ pointwise at almost every $x$.

[If you really don't have a proof of this already, then take a look at Rudin Theorem 3.12.]

Let's apply this theorem to part (1) of your problem. In your original post, you observed that, for all $n \in \mathbb N$, we can find a $g_n(x)$ and $h_n(x)$ in $L^1(\mathbb R^n)$ such that $$g_n(x) \leq f(x) \leq h_n(x) \ \ \ \ \ \ \ \ {\rm and} \ \ \ \ \ \ \ \ \ \ || g_n - h_n ||_{L^1(\mathbb R^n)} < \frac 1 n.$$ Applying the above theorem to $u_n (x) = h_n(x) - g_n(x)$, we find that there must exist subsequences $g_{n_m}(x)$ and $h_{n_m}(x)$ such that $g_{n_m} (x)$ and $h_{n_m}(x)$ both converge to $f(x)$ pointwise at almost every $x$. Hence $f(x)$, being the a.e. pointwise limit of a sequence of Lebesgue measurable functions, must also be Lebesgue measurable.

• This is a nice proof -- I forgot about the relationship between pointwise limit and $L^p$ limit. Thank you! – Yuxin Wang Aug 30 '17 at 18:55
• Regarding the edit, I think your argument is fine. You established that $u_{n_m} = h_{n_m} - g_{n_m} \to 0$ a.e. Now use the inequalities $0 \leq h_{n_m} - f \leq h_{n_m} - g_{n_m} = u_{n_m}$ to conclude that $h_{n_m} \to f$ a.e. Similarly, $g_{n_m} \to 0$ a.e. – Bungo Aug 30 '17 at 19:20
• @Bungo Good point! I was being really stupid. :) – Kenny Wong Aug 30 '17 at 19:24

Hint: Consider the supremum of a sequence of $g$'s and the infimum of a sequence of $h$'s.

• Thanks you! I will write a complete proof based on your hint. – Yuxin Wang Aug 30 '17 at 18:40

(1) Measurability of $f(x)$. Thanks to the hint of @Robert Israel

Let $g_n, h_n \in L^1$ be such that $h_n(x) \leq f(x) \leq g_n(x)$ for all $x$ and $\int g_n(x)-h_n(x)\, dx < 1/n$. Let $h(x) = \limsup h_n(x)$ and $g(x) = \liminf g_n(x)$, so $h(x) \leq f(x) \leq g(x)$ for all $x$, and $h, g$ are measurable. Moreover, by Fatou's Lemma, $\int g(x)-h(x)\, dx \leq \liminf \int g_n(x)-h_n(x)\, dx =0$, so $g(x)=h(x)$ a.e. Thus $g(x)=h(x)=f(x)$ a.e., so $f$ is measurable.

(2) $f \in L^1$. The original proof is correct, and an easier version is pointed out by @Bungo.

• Is a separate proof that $f\in L^1$ necessary if we already know that $f$ is measurable and $g\leq f\leq h \Rightarrow \lvert f\rvert\leq \max(\lvert g\rvert, \lvert h\rvert)\in L^1$ for some $g, h\in L^1$? It seems to follow directly from the hypotheses of the problem after we prove measurability. – Michael L. Aug 30 '17 at 19:26
• @MichaelLee Good point. The additional inequality $\max(|g|,|h|) \leq |g| + |h|$ may make it even clearer. – Bungo Aug 30 '17 at 20:32