Given a measure space $(\Omega, \Sigma, \mu) = (\mathbb R, \mathscr B(\mathbb R), \lambda)$

(Or: a similar probability space on some subset of $\mathbb R$ that has Lebesgue measure 1).

Let $E \in \Sigma$ s.t. $\mu(E) < \infty$.

Let $f_1, f_2, ...\in \mathscr L^{1} (\Omega, \Sigma, \mu)$ s.t. $f_i: E \to \mathbb R$.

Let $f$ be a random variable or measurable function s.t. $f: E \to \mathbb R$ and $\forall n \in \mathbb N, \forall x \in E, \ |f_n - f| < \frac{1}{n}$.


  1. $$\int_E |f| d\mu < \infty$$

  2. $$\lim \int_E f_n d\mu \to \int_E f d\mu$$

What I tried:

  1. $$\int_E |f| d\mu = \int_E |f - f_n + f_n| d\mu$$

$$\le \int_E |f - f_n| + |f_n| d\mu$$

$$\le \int_E |f - f_n| d\mu + \int_E |f_n| d\mu$$

$$\le \int_E \frac{1}{n} d\mu + \int_E |f_n| d\mu$$

$$\le \frac{1}{n} \mu(E) + \int_E |f_n| d\mu < \infty$$

  1. $$\lim \int_E |f - f_n| d\mu$$

$$\le \lim \int_E \frac{1}{n} d\mu$$

$$\le \lim \frac{1}{n} \mu(E) = 0$$

By corollary of the dominated convergence theorem, we have

$$\lim \int_E f_n d\mu \to \int_E f d\mu$$

Is that right?

  • 2
    $\begingroup$ Everything looks pretty good. For the first one, just point out at the end that $\int_E|f_n|d\mu<\infty$ since $f_n$ is $L_1$. For the second one if you want to proceed directly using dominated convergence, use the fact that $|f_n|\leq |f-f_n|+|f|\leq 1/n+|f|$, and $|f|\leq 1/n+|f_n|$ where the r.h.s. is integrable on $E$ and therefore a valid upper bound for DCT. Note that in either case, you'd need to show $f_n(x)\rightarrow f(x)$, which is trivial from your assumptions. $\endgroup$
    – Alex R.
    May 13, 2016 at 21:42
  • $\begingroup$ @AlexR. Post as answer? $\endgroup$
    – BCLC
    May 20, 2016 at 19:00

1 Answer 1


In short, both parts look correct. For the second part, to be really pedantic I think you technically should be using the $\lim\sup$ and not $\lim$ because otherwise you are already assuming a priori that the limit actually exists.

But since the limit does turn out to exist, and when the limit exists it equals the limit superior, this turns out not to matter much, to the extent I am not even certain my above objection is correct.

I am just posting the same solution as yours below, but with more details, so that you or anyone else reading this post in the future can have it as a reference. This turned out to be a longer post than I had planned, because every time I wrote out a step, I thought of another simple intermediate step which I could write out explicitly too -- thus in some cases I imagine that the lack of brevity here may hurt more than it helps, although I hope that every individual step is elementary enough by itself that this does not become an issue for those attempting to use this as a reference.

In what follows, please note that $\mathbb{1}_E$ denotes the indicator function for the measurable set $E$, i.e. $$\mathbb{1}_E (x) = \begin{cases} 1 & x \in E \\ 0 & x \not\in E \end{cases} $$

1. $$\int_E |f| d\mu = \int_E |f - f_n + f_n| d\mu \,.$$

By Triangle Inequality,

$$\le \int_E |f - f_n| + |f_n| d\mu \,.$$

By additivity of the integral,

$$= \int_E |f - f_n| d\mu + \int_E |f_n| d\mu \,.$$

By monotonicity of the integral and the assumption that $|f-f_n| < \frac{1}{n}$:

$$\le \int_E \frac{1}{n} d\mu + \int_E |f_n| d\mu \,.$$

By linearity of the integral $\int_E \frac{1}{n} d \mu = \frac{1}{n} \int_E 1 d\mu$, which then means that the above, more or less by definition of Lebesgue measure ($\int_E 1 d\mu = \int \mathbb{1}_E d \mu = \mu(E)$):

$$= \frac{1}{n} \mu(E) + \int_E |f_n| d\mu\,.$$

Because by assumption $\mu(E) < \infty$, definition of $L^1$, and assumption that all $f_n$ are $L^1$: $$< \infty\,.$$ Thus, by transitivity, $\int_E |f| d \mu < \infty$.

2. To apply the dominated convergence theorem directly, you need to make sure all of the relevant assumptions are satisfied. As Alex R. already noted above in the comments, you would need to show first that the $f_n$ approach $f$ pointwise. This should follow from your assumption plus the triangle inequality.

But then I don't think the assumptions give you (at least not directly) a dominating function for the $f_n$, so you cannot apply the dominated convergence theorem to the sequence $f_n$ directly.

However, you can apply the dominated convergence theorem to the sequence of functions $f_n - f$, since your assumption implies that $\lim (f_n -f)(x) = 0$ for all $x \in E$, thus pointwise $f_n -f \to 0$, and the assumption also gives the dominating function itself, namely $1$ (i.e. $|f_n - f| \le 1$ for all $n$).

Note that $1$ only works as a dominating function here because of our assumption that $\mu(E):=\int_E 1 d\mu < \infty$; technically it would probably be more correct to say that $\mathbb{1}_E$ is the dominating function, rather than $1$ itself.

As a result of applying the dominated convergence theorem to this sequence of functions, we get the desired result, namely that $$\lim \int_E f_n - f d\mu = \int_E 0 d \mu =0 \implies \lim \int_E f_n d \mu = \lim \int_E f d \mu = \int_E f d \mu $$ by additivity of the integral and the fact that the limit of a constant sequence is just the constant itself.

Note: technically, we are applying the dominated convergence theorem (for measurable functions on $\mathbb{R}$) to the sequence of functions $f_n \mathbb{1}_E - f \mathbb{1}_E = \mathbb{1}_E(f_n - f)$, but this makes no difference, since a function $h$ is measurable on $E$ if and only if it equals $\mathbb{1}_E g$ for some measurable function $g: \mathbb{R} \to \mathbb{R}$.

What follows below is in some sense a rehash of the proof of the dominated convergence theorem restricted to the special case under consideration in this problem, and is essentially an elaboration on your solution above, albeit with limit superiors used instead of limits at several places.

Since for all $n$, $|f-f_n| \ge 0$ by definition of absolute value, we have for all $n$ by monotonicity of the integral (and fact that $\mathbb{1}_E \ge 0$), that $\int_E |f_n -f| d \mu \ge 0$.

Thus $\lim\inf \int_E |f_n - f| d\mu \ge 0$ (i.e. $0$ is a lower bound, and $\inf$ means the greatest lower bound).

Now as you pointed out above, $$ \lim\sup \int_E |f_n - f| d \mu \le \lim\sup \int_E \frac{1}{n} d \mu $$ by monotonicity of the integral ($|f_n - f| < \frac{1}{n}$ by assumption ) and the fact that we are taking upper bounds and the supremum is the least upper bound.

By linearity of the integral, the above $$ = \lim\sup \frac{1}{n} \int_E 1 d \mu = \lim\sup \frac{1}{n} \mu(E) = \mu(E) \lim\sup \frac{1}{n} $$ following by definition of $\mu(E)$, and linearity of limit and thus homogeneity limit superior for non-negative constants (i.e. note that by definition of measure that $\mu(E) \ge 0$).

Anyway, we know that $\lim \frac{1}{n}$ exists and equals $0$, and when the limit exists it equals the limit superior, so $\lim\sup \frac{1}{n} = 0$. Since $\mu(E) < \infty$, we can thus conclude that: $$\mu(E) \lim\sup \frac{1}{n} = \mu(E) (0) = 0.$$

In particular, by transitivity, we have shown that: $$\lim\sup \int_E |f_n - f| d \mu \le 0 ,$$ and further that $$\lim\sup \int_E |f_n - f| d \mu \le 0 \le \lim\inf \int_E |f_n - f| d \mu \\ \implies \lim\sup \int_E |f_n - f| d \mu \le \lim\inf \int_E |f_n - f| d \mu. $$ Since in general it holds that the limit superior of a sequence is greater than or equal to the limit inferior of a sequence, we have shown that $$\lim\sup \int_E |f_n - f | d \mu = \lim\inf \int_E |f_n - f| d \mu. $$ Thus as a result, $$0 \le \lim\sup \int_E |f_n - f| d \mu \le 0 \quad \text{or} \quad 0 \le \lim\inf \int_E |f_n -f| d \mu \le 0$$ i.e. in particular, $$ \lim \int_E |f_n - f| d \mu =0.$$ (Since the limit exists if and only if the limit superior equals the limit inferior, in which case it is equal to both.) The idea of the above is essentially analogous to that of the Squeeze/Sandwich theorem from more elementary calculus.

Anyway, the fact that $\lim \int_E |f_n - f| d \mu = 0$ implies fairly readily the desired result that $\lim \int_E f_n d \mu = \lim \int_E f d \mu$, as can be seen in the proof contained in the Wikipedia article for the dominated convergence theorem -- I will now hash out the details of this implication, largely following the Wikipedia argument, but also with some more detail.

Now, by the definition of absolute value, we have immediately that for all $n$, $$\left| \int_E f_n - f d \mu \right| \ge 0\, $$ thus $$\lim\inf \left| \int_E f_n - f d \mu \right| \ge 0, $$ remembering that infimum is the greatest lower bound.

We have, using the fact $|\int g d\mu| \le \int | g | d \mu$ (see here) we get that: $$\lim\sup \left| \int_E f_n -f d \mu \right| \le \lim \int_E | f_n - f | d \mu \le 0 $$ using what we have shown above and the fact that the supremum is the least upper bound.

Completely analogous to the above again, we get that $$\lim\sup \left| \int_E f_n -f d \mu \right| \le 0 \le \lim\inf \left| \int_E f_n - f d \mu \right| \\ \implies \lim \left| \int_E f_n -f d \mu \right| \text{ exists, and }\lim\left| \int_E f_n - f d \mu \right| =0.$$

Then the desired result follows from the facts that by linearity of the integral: $$\lim \left| \int_E f_n d\mu - \int_E f d\mu \right| =\lim \left| \int_E f_n -f d \mu \right| = 0 $$ and that $$ \lim \left| \int_E f_n d\mu - \int_E f d\mu \right|= 0 \implies \lim \left[ \int_E f_n d\mu - \int_E f d\mu \right] = 0 \\ \implies \lim \int_E f_n d \mu - \lim \int_E f d\mu =0 $$ using linearity of limits, $$\implies \lim \int_E f_n d \mu = \lim \int_E f d\mu = \int_E f d\mu, $$ since the limit of a constant sequence is the constant itself.

  • $\begingroup$ Thanks William! ^-^ No offense but I think some parts of your answer are kind of spoonfeeding. It's good that you emphasise, but I think you may just leave some of the emphasis to the reader by replacing details with something like '(Why?)' $\endgroup$
    – BCLC
    Dec 20, 2016 at 13:48
  • 1
    $\begingroup$ @BCLC Yeah I definitely agree with the spoonfeeding -- to be honest writing out this answer was as much for my own benefit (i.e. in terms of rehashing material I have not worked with in a long time and testing myself) as for any other reason. In an ideal world it should be pared down a lot more. $\endgroup$ Dec 20, 2016 at 14:11

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