# Let $f:\mathbb{R}\to[0,\infty)$ measurable and $f\in L^1$. Show that $\mu(E)<\delta \implies \int_E f < \varepsilon$.

I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help to understand the solution to the following problem:

Let $$f:\mathbb{R}\to[0,\infty)$$ measurable and $$f\in L^1$$. Show that for every $$\varepsilon>0$$, there exists a $$\delta>0$$ such that

$$E\in\mathcal{M},\; \mu(E)<\delta \implies \int_E f < \varepsilon.$$

(A quick note: $$\mathcal{M}$$ is the set of measurable subsets of $$\mathbb{R}$$ and the set function $$\mu : \mathcal{M} \rightarrow [0, +\infty]$$ is the Lebesgue measure).

Solution: For every $$n\in\mathbb{N}$$ consider the function $$f_n(x)=\min\{f(x),n\}$$. Note that $$f_n \leq n$$. From the Monotone Convergence Theorem $$\color{red}{(1)}$$ we have that

$$\lim_{n\to\infty}\int f_n = \int f.$$

Let $$\varepsilon>0$$. We can find $$n\in\mathbb{N}$$ such that

$$\int (f-f_n) = \int f - \int f_n < \frac{\varepsilon}{2}\;\;\;\;\;\color{red}{(2)}.$$

We choose $$\delta = \frac{\varepsilon}{2n}$$. Let $$E \subseteq \mathbb{R}$$ with $$\mu(E)<\delta$$. It follows that

$$\int_E f = \int_E f_n + \int_E (f-f_n) \leq \int_E f_n + \int (f-f_n) \leq \underbrace{n\mu(E)}_{\color{red}{(3)}} + \frac{\varepsilon}{2} < n\frac{\varepsilon}{2n} + \frac{\varepsilon}{2} = \varepsilon.$$

I marked in red the parts of the solution that I don't understand, specifically

$$\color{red}{(1)}$$ For the Monotone Convergence Theorem to apply the conditions are that $$\{f_n\}$$ is nondecreasing and that $$f_n$$ converges pointwise to $$f$$. I don't see why both of these conditions hold here.

$$\color{red}{(2)}$$ The way that I interpret $$\color{red}{(2)}$$ is that it is possible to find $$n$$ sufficiently large such that $$f$$ and $$f_n$$ differ from a value that is less than $$\frac{\varepsilon}{2}$$. Is that the correct interpretation?

$$\color{red}{(3)}$$ I don't understand why the (Lebesgue) integral $$\int_E f_n$$ is bounded by $$n\mu(E)$$.

Looking back at the problem as a hole it is also unclear to me why we had to define $$f_n(x)=\min\{f(x),n\}$$.

For 1. $n<n+1$ so $\min\{f(x),n\}<\min\{f(x),n+1\}$. Thus $f_n(x)<f_{n+1}(x)$. Also by the Archimedian property, for all $f(x)$ there is some $n$ where $n>f(x)$. Thus for all $x$, there is some $n$ where $\min\{f(x),n\}=f(x)$.

For 2. You're just using the definition of convergence. $\int f_n$ is just a sequence of numbers converging to $\int f$, another number. So we can find some $n$ where $|\int f-\int f_n|<\frac{\epsilon}{2}$. Noting that $\int f>\int f_n$ we can remove the absolute value.

For 3. For all $x$, $f_n(x)\leq n$. So $\int_E f_n(x) dx \leq \int_E n dx=n\mu(E)$. This is the reason for defining $f_n(x)$.

• I couldn't hope for a better answer! Thank you for your detailed and clear explanation. It is extremely helpful and understandable. May 6, 2016 at 20:30
1) $f_n(x)=\min(f(x),n)$ is non decreasing from basic properties of $min$. To better see this, prove this fact for each fixed $x$. Also $f_n(x)\rightarrow f(x)$ since $f(x)$ is assumed to be finite almost everywhere (otherwise it wouldn't be $L^1$).
2) No, this is a directly consequence of monotone convergence theorem and the definition of limits. The integrals are within $\epsilon/2$.
3) Look at the definition of $f_n$. Clearly $f_n\leq n$.
• "otherwise it wouldn't be $L^1$". Well more importantly the theorem says $f\colon \Bbb{R} \to [0,\infty)$