$f_n \rightarrow f$ & $|f_n|\le g\in L_1$ Prove: $f\in L_1$ | $\lim_{n\rightarrow\infty} \int_X f_n d\mu=\int_X f d\mu$ | $f_n\rightarrow f$ in $L_1$

Question

Let $(f_n)$ be a sequence of measurable functions on $X$, for which there exist such $f, g$, that $f_n \rightarrow f$ and $\forall_n \: |f_n| \le g$ for $g \in L_1 (X, \mu) \Leftrightarrow \int_X |g| d\mu < \infty$

Prove, that

1. $f \in L_1 (X, \mu)$
2. $\lim_{n \rightarrow \infty} \int_X f_n d\mu = \int_X f d\mu$
3. $f_n \rightarrow f$ in $L_1 (X, \mu)$

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My attempt

1. If $\forall_n \: |f_n| \le g$, and we know, that $f_n \rightarrow f$, then it's also true that $|f| \le g$. Thus:

$$|f| \le g \quad \Leftrightarrow \quad |f| \le |g| \quad \Leftrightarrow \quad \int_X |f| d\mu \le \int_X |g| d\mu < \infty$$

2. It's basically the Dominated convergence theorem

3. $f_n \rightarrow f$ means (I think!), that $h_n = f_n - f$ is an $L_1 (X, \mu)$ function and that $\lim_{n \rightarrow \infty} \int_X h_n d\mu = 0$. The latter is a consequence of 2. (just subtract the one side from the other), we prove the former:

$$|h_n| = |f_n - f| \le |f_n| + |f| \le |g| + |f| \quad \Leftrightarrow \quad \int_X |h_n| d\mu \le \int_X |g| d\mu + \int_X |f| d\mu < \infty \text{ (by definition) }+ \infty \text{ (by 1.) } < \infty$$

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I am not really sure about 1., if we can just assume that "if $|f_n| \le |g|$, then $|f| \le |g|$".

Also, I don't know if we cannot simply subtract the one side by the other in point 2 and simply call it a day, as because $f_n \rightarrow f \quad \Leftrightarrow \quad f_n - f \rightarrow 0$

Point 3 seems to be a consequence from the other two points, so I guess it should be fine? Unless I absolutely misunderstood what "convergence in $L_1 (X, \mu)$" means

Answer 1

Ignore my previous comments; they probably made it sound more complicated than it was. Your attempts are almost correct. Regarding your concern about point (1), that is a basic fact about limits:

Suppose $\{a_n\}$ and $\{b_n\}$ are sequences in $[-\infty,\infty]$ which converge in $[-\infty,\infty]$, and suppose that for all $n\in\Bbb{N}$, we have $a_n\leq b_n$. Then, $\lim\limits_{n\to\infty}a_n\leq \lim\limits_{n\to\infty}b_n$.

This is why $|f_n|\leq g$ for all $n$ implies $|f|=|\lim f_n|=\lim|f_n|\leq \lim g=g$, and hence $\int|f|\leq \int g<\infty$.

For 3 I think you have the right idea but your presentation isn't really clear. Just write things in words, and avoid unnecessary symbols. Saying $f_n\to f$ in $L^1$ means you have to show $\lim\limits_{n\to\infty}\int|f_n-f|=0$ (you were missing absolute values). To prove this, note that $f_n\to f$ pointwise by hypothesis and hence $|f_n-f|$ converges to $0$ pointwise, and $|f_n-f|\leq |f_n|+|f|\leq 2g$, and $2g\in L^1$, so by DCT (the version you seem to know by heart), $\lim\limits_{n\to\infty}\int|f_n-f|=0$.

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In my comment, my intention was to merely point out that sometimes, the DCT is simply stated as

If $\{f_n\}$ is a sequence of measurable functions which converges pointwise to a function $f$ and $g\in L^1(\mu)$ is a function such that for all $n$, $|f_n|\leq g$ pointwise, then $\lim\limits_{n\to\infty}\int_X|f_n-f|=0$.

From this minimalistic way of phrasing things, there are already several things one can observe.

• It is implicit in the statement (but also the only logical consequence of the hypothesis) that $g$ is a non-negative function.
• $f$ being a pointwise limit of measurable functions is also measurable (this is a basic fact from measure theory).
• $|f_n|\leq g$ and $g\in L^1$ implies that each $f_n\in L^1$.
• To prove $f\in L^1$, we can argue as you have done, or we can note the following: $\int|f|\leq \int|f-f_n|+\int|f_n|\leq \int|f_n-f|+\int|g|$, and on the RHS, the terms converge to $0$, so in particular they are finite.
• Finally, we can conclude $\lim\limits_{n\to\infty}\int f_n=\int f$ because of the very simple estimate $\left|\int f_n-\int f\right|=\left|\int (f_n-f)\right|\leq \int|f_n-f|$, and this converges to $0$ by assumption. So by the squeeze theorem, it follows that $\lim\limits_{n\to\infty}\int f_n=\int f$.

Answer 2

Theorem. If $f_n\to f$ $\mu$-a.e. and $|f_n|\le g$ with $g\in L^1(\mu)$, then $\int f_n\,\mathrm d\mu\to\int f\,\mathrm d\mu$.

Suppose the conditions of Theorem hold. We show (3).

• Apply Theorem to the sequence $|f_n|$ (with absolute values): we have $|f_n|\to|f|$ $\mu$-a.e. (by continuous mapping) and $\Bigl||f_n|\Bigr|=|f_n|\le g$, so by the theorem $\int_X|f_n|\,\mathrm d\mu\to\int_X|f|\,\mathrm d\mu$.
• To conclude that $f_n\to f$ in $L^1(\mu)$, use the Riesz-Scheffé lemma which is a direct consequence of Fatou's lemma: \begin{align*} 2\int|f|\,\mathrm d\mu-\limsup_{n\to\infty}\int|f_n-f|\,\mathrm d\mu &=\liminf_{n\to\infty}\int\underbrace{(|f_n|+|f|-|f_n-f|)}_{\ge0}\,\mathrm d\mu\\&\underset{\text{Fatou}}\ge\int\liminf_{n\to\infty}\:(|f_n|+|f|-|f_n-f|)\,\mathrm d\mu\\[.4em]&=2\int|f|\,\mathrm d\mu, \end{align*} so $$\lim_{n\to\infty}\int|f_n-f|\,\mathrm d\mu=0.$$