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 $f_n(x)=\min\{f(x),n\}$.


2 Answers 2


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 $f_n(x)$.

  • $\begingroup$ I couldn't hope for a better answer! Thank you for your detailed and clear explanation. It is extremely helpful and understandable. $\endgroup$
    – glpsx
    May 6, 2016 at 20:30
  • 1
    $\begingroup$ @VonKar No worries. I'm glad you found it helpful. $\endgroup$
    – user223391
    May 6, 2016 at 20:34

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 $f_n\leq n$.

  • $\begingroup$ "otherwise it wouldn't be $L^1$". Well more importantly the theorem says $f\colon \Bbb{R} \to [0,\infty)$ $\endgroup$
    – user223391
    May 6, 2016 at 20:24

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