Notions of convergence of measurable functions

A sequence of measurable functions can converge to a measurable function in several senses.

Notions of convergence#

Definition 1 (Notions of convergence). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, and let ${f_n \colon X \to \mathbb{R},}$ for every ${n,}$ and ${f \colon X \to \mathbb{R}}$ be measurable.
(i) We say that ${f_n \to f}$ pointwise if ${\lim_{n \to \infty} f_n(x) = f(x)}$ for every ${x \in X.}$
(ii) We say that ${f_n \to f}$ uniformly if ${\lim_{n \to +\infty} \sup_{x \in X} |f_n(x) - f(x)| = 0.}$
(iii) We say that ${f_n \to f}$ almost everywhere if ${\lim_{n \to +\infty} f_n(x) = f(x)}$ for almost every ${x \in X.}$

(iv) Let ${p \in [1, +\infty].}$ We say that ${f_n \to f}$ in $L^p(X)$ if ${\lim_{n \to +\infty} \|f_n - f\|_p = 0,}$ that is, if

$$\begin{aligned} &\lim_{n \to \infty} \int_X |f_n - f|^p \, d\mu = 0 && \text{when } p < +\infty, \\ &\lim_{n \to \infty} \operatorname*{ess\,sup}_{x \in X} |f_n(x) - f(x)| = 0 && \text{when } p = +\infty. \end{aligned}$$

(v) We say that ${f_n \to f}$ in measure if, for every ${\varepsilon > 0,}$

$$\lim_{n \to \infty} \mu(\{ x \in X : |f_n(x) - f(x)| > \varepsilon \}) = 0.$$

The set in Definition 1 (v) is also written ${\{ |f_n - f| > \varepsilon \}.}$

Remark 2. The notions (iii), (iv) and (v) of Definition 1 make sense for ${(f_n) \subset L^p}$ and ${f \in L^p.}$

Remark 3 (Obvious implications). Let $(f_n)$ and $f$ be as in Definition 1. The notions of convergence of Definition 1 are related as in Figure 1. Uniform convergence implies pointwise convergence, which implies convergence almost everywhere, and neither implication can be reversed. Uniform convergence implies convergence in ${L^\infty,}$ which does not imply uniform convergence, and convergence in $L^\infty$ implies convergence almost everywhere and convergence in measure.

Convergence in $L^\infty$ implies convergence almost everywhere: ${|f_n(x) - f(x)| \le \|f_n - f\|_\infty}$ for almost every ${x \in X,}$ and ${\lim_{n \to \infty} \|f_n - f\|_\infty = 0.}$ Convergence in $L^\infty$ implies convergence in measure. Let ${\varepsilon > 0}$ and consider the set ${\{ |f_n - f| > \varepsilon \}.}$ There exists $N$ such that ${\|f_n - f\|_\infty \le \varepsilon/2}$ for every ${n \ge N.}$ Hence ${\mu(\{ |f_n - f| > \varepsilon \}) = 0}$ for every ${n \ge N.}$

Figure 1. The drawing of Remark 3: the five notions of convergence of $f_n$ to ${f,}$ a double arrow for each implication and a crossed double arrow for each implication that does not hold.

Typewriter sequence#

Example 4 (Typewriter sequence). On ${[0, 1]}$ with the Lebesgue measure ${\lambda,}$ consider the sequence ${(f_n)_{n \ge 1}}$ of characteristic functions drawn in Figure 2. Its first terms are ${f_1 = \chi_{[0, 1/2]}}$ and ${f_2 = \chi_{[1/2, 1]},}$ then the characteristic functions ${f_3, \dots, f_6}$ of ${[0, 1/4],}$ ${[1/4, 1/2],}$ $[1/2, 3/4]$ and ${[3/4, 1],}$ then ${f_7 = \chi_{[0, 1/8]},}$ and so on. In detail, the terms come in groups: for every ${k \ge 1}$ the $k$-th group is made of the characteristic functions of the intervals ${[j/2^k, (j + 1)/2^k]}$ with ${j = 0, 1, \dots, 2^k - 1,}$ in this order.

Figure 2. The drawing of Example 4: the graphs of the functions $f_1$ to ${f_{14},}$ in rows of 2, 4 and 8, each on its own copy of ${[0, 1].}$ The functions ${f_1,}$ $f_3$ and ${f_7,}$ circled, are the first terms of the subsequence that converges to $0$ almost everywhere.

For every ${x \in [0, 1],}$ ${\varliminf_{n \to \infty} f_n(x) = 0}$ and ${\varlimsup_{n \to \infty} f_n(x) = 1.}$ In fact, for every $x$ and every $N$ there exist ${n_1, n_2 \ge N}$ such that ${f_{n_1}(x) = 0}$ and ${f_{n_2}(x) = 1.}$ Indeed, for every ${k \ge 2}$ the point $x$ lies in at least one and in at most two of the intervals of the $k$-th group. Hence the limit ${\lim_{n \to \infty} f_n(x)}$ exists at no ${x,}$ and $(f_n)$ does not converge almost everywhere.

The circled functions of Figure 2 form a subsequence ${(f_{n_k}),}$ made of the characteristic functions of ${[0, 1/2],}$ ${[0, 1/4],}$ ${[0, 1/8],}$ and so on, such that ${f_{n_k} \to 0}$ almost everywhere on ${[0, 1].}$ Finally, ${f_n \to 0}$ in ${L^1([0, 1]).}$

Convergence in measure#

Theorem 5 (Almost everywhere versus in measure). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, and let ${f_n \colon X \to \mathbb{R},}$ for every ${n,}$ and ${f \colon X \to \mathbb{R}}$ be measurable.
(1) If ${\mu(X) < +\infty}$ and ${f_n \to f}$ almost everywhere, then ${f_n \to f}$ in measure.
(2) If ${f_n \to f}$ in measure, then there exists a subsequence $(f_{n_k})$ such that ${f_{n_k} \to f}$ almost everywhere.

Theorem 5 (2) gives the convergence almost everywhere of a subsequence: the typewriter sequence of Example 4 converges to $0$ in measure, but $(f_n)$ itself does not converge almost everywhere. In Theorem 5 (1) the assumption ${\mu(X) < +\infty}$ is needed. Otherwise, take ${(\mathbb{R}, \mathcal{L}(\mathbb{R}), \lambda)}$ and ${f_n = \chi_{[n, n+1]},}$ as in Figure 3: then ${f_n \to 0}$ almost everywhere, but $(f_n)$ does not converge to $0$ in measure. In fact, ${\lambda(\{ |f_n| > 1/2 \}) = \lambda([n, n + 1]) = 1}$ for every ${n.}$

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Figure 3. The function ${f_n = \chi_{[n, n+1]}}$ on ${\mathbb{R},}$ the dashed level ${1/2,}$ and the set ${\{ |f_n| > 1/2 \} = [n, n + 1]}$ hatched on the axis. The slider sets ${n.}$

Theorem 6 ($L^p$ implies measure). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${f_n \colon X \to \mathbb{R},}$ for every ${n,}$ and ${f \colon X \to \mathbb{R}}$ be measurable, and let ${p \in [1, +\infty].}$ If ${f_n \to f}$ in ${L^p,}$ then ${f_n \to f}$ in measure. In particular, by Theorem 5 (2), there exists a subsequence $(f_{n_k})$ such that ${f_{n_k} \to f}$ almost everywhere.

Proof (complete). The case ${p = +\infty}$ is in Remark 3. Let ${p \in [1, +\infty)}$ and let ${\varepsilon > 0.}$ Then

$$\begin{aligned} \biggl( \int_X |f_n - f|^p \, d\mu \biggr)^{\frac{1}{p}} &\ge \biggl( \int_{\{ |f_n - f| > \varepsilon \}} |f_n - f|^p \, d\mu \biggr)^{\frac{1}{p}} \\ &\ge \biggl( \int_{\{ |f_n - f| > \varepsilon \}} \varepsilon^p \, d\mu \biggr)^{\frac{1}{p}} \\ &= \big( \varepsilon^p \, \mu(\{ |f_n - f| > \varepsilon \}) \big)^{\frac{1}{p}} \\ &= \varepsilon \, \mu(\{ |f_n - f| > \varepsilon \})^{\frac{1}{p}}, \end{aligned}$$

where both inequalities are monotonicity, (3) and (2) of Remark 4 of the fifth post. Hence

$$\mu(\{ |f_n - f| > \varepsilon \}) \le \Big( \frac{1}{\varepsilon} \Big)^p \|f_n - f\|_p^p,$$

where ${\|f_n - f\|_p^p \to 0}$ as ${n \to +\infty}$ by assumption. Therefore

$$\lim_{n \to \infty} \mu(\{ |f_n - f| > \varepsilon \}) = 0. \tag*{$\blacksquare$}$$

Remark 7. On ${(\mathbb{R}, \mathcal{L}(\mathbb{R}), \lambda)}$ let ${f_n = n \chi_{[0, 1/n]}}$ for every ${n \ge 1,}$ as in Figure 4. Then ${f_n \to 0}$ almost everywhere, ${f_n \nrightarrow 0}$ in ${L^1,}$ and ${f_n \to 0}$ in measure.

In fact, ${\int_{\mathbb{R}} |f_n| \, d\lambda = 1}$ for every ${n.}$ Moreover, for every ${\varepsilon > 0}$ we have ${\{ |f_n| > \varepsilon \} = [0, 1/n]}$ for every $n$ sufficiently big, hence ${\lambda(\{ |f_n| > \varepsilon \}) = 1/n \to 0.}$

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ε

1.50

Figure 4. The drawing of Remark 7: the function ${f_n = n \chi_{[0, 1/n]},}$ the region of area $1$ under its graph, the dashed level ${\varepsilon,}$ and the set ${\{ |f_n| > \varepsilon \}}$ hatched on the axis. The sliders set $n$ and ${\varepsilon.}$

Convergence in $L^p$#

Theorem 8 (Dominated convergence in $L^p$). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${f_n \colon X \to \mathbb{R},}$ for every ${n,}$ and ${f \colon X \to \mathbb{R}}$ be measurable, and let ${p \in [1, +\infty).}$ Suppose that ${f_n \to f}$ almost everywhere and that there exists ${g \in L^p}$ such that ${|f_n| \le g}$ almost everywhere for every ${n.}$ Then ${f \in L^p}$ and ${f_n \to f}$ in ${L^p.}$

Remark 9. Theorem 8 does not hold with ${p = +\infty.}$ On ${(\mathbb{R}, \mathcal{L}(\mathbb{R}), \lambda)}$ let ${f_n = \chi_{[0, 1 - 1/n]}}$ for every ${n \ge 1,}$ as in Figure 5. Then ${f_n \to \chi_{[0, 1]}}$ almost everywhere and ${|f_n| \le 1 \in L^\infty,}$ but ${f_n \nrightarrow \chi_{[0, 1]}}$ in ${L^\infty.}$

In fact, ${|\chi_{[0, 1]}(x) - f_n(x)| = \chi_{(1 - 1/n, 1]}(x)}$ for every ${x \in \mathbb{R},}$ hence

$$\|f_n - \chi_{[0, 1]}\|_\infty = \|\chi_{(1 - 1/n, 1]}\|_\infty = 1 \qquad \text{for every } n,$$

so ${\|f_n - \chi_{[0, 1]}\|_\infty}$ does not tend to $0$ as ${n \to \infty.}$

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Figure 5. The drawing of Remark 9: above, the function ${f_n = \chi_{[0, 1 - 1/n]}}$ and, dashed, its limit ${\chi_{[0, 1]};}$ below, their difference ${|\chi_{[0, 1]} - f_n| = \chi_{(1 - 1/n, 1]}.}$ The slider sets ${n.}$

Remark 10 (Reverse dominated convergence in $L^p$). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${f_n, f \colon X \to \mathbb{R}}$ be measurable, for every ${n.}$ Let ${p \in [1, +\infty]}$ and ${(f_n) \subset L^p.}$ If ${f_n \to f}$ in ${L^p,}$ then ${f \in L^p}$ and there exist a subsequence $(f_{n_k})$ and ${g \in L^p}$ such that:
(i) ${f_{n_k} \to f}$ almost everywhere;
(ii) ${|f_{n_k}| \le g}$ almost everywhere for every ${k.}$

Proposition 11 (Almost everywhere and norms imply $L^p$). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${f_n, f \colon X \to \mathbb{R}}$ be measurable, for every ${n,}$ and let ${p \in [1, +\infty).}$ Suppose that ${f_n \to f}$ almost everywhere, that ${f_n, f \in L^p}$ and that ${\lim_{n \to \infty} \|f_n\|_p = \|f\|_p.}$ Then ${f_n \to f}$ in ${L^p.}$

Proof (complete). For every ${x \in X,}$

$$|f_n(x) - f(x)|^p \le 2^{p-1} \big( |f_n(x)|^p + |f(x)|^p \big).$$

Hence the functions

$$g_n := 2^{p-1} \big( |f_n|^p + |f|^p \big) - |f_n - f|^p$$

satisfy ${g_n \ge 0,}$ and at almost every ${x \in X}$

$$\lim_{n \to +\infty} g_n(x) = 2^{p-1} \big( |f(x)|^p + |f(x)|^p \big) - 0 = 2^p |f(x)|^p.$$

By Fatou’s lemma, Theorem 1 of the previous post,

$$\begin{aligned} 2^p \|f\|_p^p = \int_X 2^p |f|^p \, d\mu &\le \varliminf_n \int_X g_n \, d\mu \\ &= \varliminf_n \int_X \Big( 2^{p-1} \big( |f_n|^p + |f|^p \big) - |f_n - f|^p \Big) \, d\mu \\ &= \varliminf_n \Big( 2^{p-1} \|f_n\|_p^p + 2^{p-1} \|f\|_p^p - \|f_n - f\|_p^p \Big). \end{aligned}$$

The last expression equals

$$\begin{aligned} &\bigl( \varliminf_n 2^{p-1} \|f_n\|_p^p \bigr) + 2^{p-1} \|f\|_p^p - \varlimsup_n \|f_n - f\|_p^p \\ &\quad = 2^{p-1} \|f\|_p^p + 2^{p-1} \|f\|_p^p - \varlimsup_n \|f_n - f\|_p^p \\ &\quad = 2^p \|f\|_p^p - \varlimsup_n \|f_n - f\|_p^p, \end{aligned}$$

where the second equality is the hypothesis ${\lim_n \|f_n\|_p = \|f\|_p.}$ Indeed, the first equality holds because, by the same hypothesis, the sequence ${2^{p-1} \|f_n\|_p^p}$ converges.

Comparing the first line of the first display with the last line of the second, ${\varlimsup_n \|f_n - f\|_p^p \le 0.}$ Indeed, ${2^p \|f\|_p^p}$ is finite because ${f \in L^p,}$ so it can be subtracted from both sides. Since ${\|f_n - f\|_p^p \ge 0}$ for every ${n,}$ it follows that ${\lim_n \|f_n - f\|_p = 0.}$ $\blacksquare$

Uniqueness of limits#

Remark 12 (Uniqueness of limits). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${f_n, f, g \colon X \to \mathbb{R}}$ be measurable, for every ${n.}$ Suppose that ${f_n \to f}$ with respect to a notion of convergence $N_1$ and ${f_n \to g}$ with respect to a notion of convergence ${N_2,}$ both among the notions of Definition 1. Then ${f = g}$ almost everywhere.

Proof (complete). Every notion of Definition 1 implies the convergence almost everywhere of a subsequence, so there exists a subsequence $(f_{n_k})$ such that ${f_{n_k} \to f}$ almost everywhere. Indeed, uniform and pointwise convergence imply convergence almost everywhere by Remark 3, and convergence in $L^p$ or in measure gives a subsequence that converges almost everywhere by Theorem 6 and by Theorem 5 (2).

Also ${f_n \to g}$ with respect to ${N_2,}$ hence ${f_{n_k} \to g}$ with respect to ${N_2,}$ and therefore there exists a subsequence $(f_{n_{k_l}})$ of $(f_{n_k})$ such that ${f_{n_{k_l}} \to g}$ almost everywhere. Moreover, ${f_{n_k} \to f}$ almost everywhere, hence ${f_{n_{k_l}} \to f}$ almost everywhere. Therefore, at almost every ${x \in X,}$

$$|f(x) - g(x)| = \lim_{l \to \infty} |f_{n_{k_l}}(x) - f_{n_{k_l}}(x)| = 0.$$

Indeed, at such $x$ the numbers $f_{n_{k_l}}(x)$ tend both to $f(x)$ and to ${g(x).}$ Hence ${f = g}$ almost everywhere. $\blacksquare$

Exercises#

The sequence $\cos(x^n)$#

Example 13. Let ${f_n(x) = \cos(x^n)}$ for ${x \in [0, 1],}$ with the Lebesgue measure $\lambda$ on ${[0, 1],}$ as in Figure 6. Does $(f_n)$ converge almost everywhere, in ${L^1,}$ in measure?

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Figure 6. The drawing of Example 13 for the $n$ of the slider: the graph of ${x^n,}$ which ends at the point ${(1, 1),}$ the graph of ${f_n(x) = \cos(x^n),}$ which ends at ${(1, \cos(1)),}$ and, dashed, the constant ${1.}$

As ${n \to \infty,}$ ${x^n \to 0}$ for ${x \in [0, 1)}$ and ${x^n \to 1}$ for ${x = 1,}$ pointwise. Hence, pointwise for every ${x \in [0, 1],}$

$$f_n(x) = \cos(x^n) \xrightarrow[n \to \infty]{} \begin{cases} 1 & \text{if } x \in [0, 1), \\ \cos(1) & \text{if } x = 1. \end{cases}$$

Therefore ${f_n \to f}$ almost everywhere, where ${f(x) = 1}$ for every ${x \in [0, 1].}$ Moreover, ${|f_n| = |\cos(x^n)| \le 1}$ for every ${x \in [0, 1].}$ Hence ${|f_n| \le f}$ for every $x$ and ${f \in L^1((0, 1)).}$ By the dominated convergence theorem, Theorem 8 with ${p = 1,}$ ${f_n \to f}$ in ${L^1((0, 1)).}$ Finally, ${f_n \to f}$ in measure by Theorem 6.

The sequence $n^{-\alpha} \chi_{[0, n]}$#

Example 14. Let ${\alpha > 0}$ and, on ${(\mathbb{R}, \mathcal{L}(\mathbb{R}), \lambda),}$ let ${f_n = (1/n^\alpha) \chi_{[0, n]}}$ for every ${n \ge 1,}$ as in Figure 7. Find all ${\alpha > 0}$ such that $(f_n)$ converges in ${L^1(\mathbb{R}),}$ and all ${\alpha > 0}$ such that $(f_n)$ converges in measure.

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0.50

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Figure 7. The drawing of Example 14: the function ${f_n = (1/n^\alpha) \chi_{[0, n]},}$ the region under its graph, of area ${n^{1 - \alpha},}$ and the dashed level ${\varepsilon,}$ with the set ${\{ f_n > \varepsilon \}}$ hatched on the axis when it is not empty. The sliders set ${n,}$ $\alpha$ and ${\varepsilon.}$

For every ${x \in \mathbb{R}}$ we have ${|f_n(x)| \le 1/n^\alpha,}$ which tends to $0$ as $n$ grows, so ${f_n \to 0}$ pointwise on ${\mathbb{R}.}$ By the uniqueness of limits, Remark 12, if $(f_n)$ converges in ${L^1,}$ it converges to $0$ in ${L^1.}$ We compute

$$\begin{aligned} \|f_n - 0\|_{L^1(\mathbb{R})} &= \int_{\mathbb{R}} |f_n| \, d\lambda \\ &= \int_{\mathbb{R}} \frac{1}{n^\alpha} \chi_{[0, n]} \, d\lambda = \frac{1}{n^\alpha} \int_0^n d\lambda \\ &= \frac{1}{n^\alpha} \, n = n^{1 - \alpha}. \end{aligned}$$

Notice that ${\lim_{n \to \infty} \|f_n\|_{L^1} = \lim_{n \to \infty} n^{1 - \alpha},}$ which is $0$ if and only if ${1 - \alpha < 0,}$ that is, if and only if ${\alpha > 1.}$ Hence $(f_n)$ converges in $L^1(\mathbb{R})$ if and only if ${\alpha > 1.}$

Let ${\varepsilon > 0.}$ By the uniqueness of limits, if $(f_n)$ converges in measure, it converges to $0$ in measure, so it is enough to study ${\lambda(\{ |f_n - 0| > \varepsilon \}) = \lambda(\{ f_n > \varepsilon \}).}$ There exists $N_\varepsilon$ such that ${f_n \le \varepsilon}$ for every ${n \ge N_\varepsilon,}$ because ${f_n \le 1/n^\alpha \to 0.}$ Hence ${\{ f_n > \varepsilon \} = \emptyset}$ for every ${n \ge N_\varepsilon.}$

Therefore ${\lambda(\{ f_n > \varepsilon \}) = \lambda(\emptyset) = 0}$ for every ${n \ge N_\varepsilon,}$ and ${\lim_n \lambda(\{ f_n > \varepsilon \}) = 0.}$ This holds for every ${\alpha > 0,}$ hence ${f_n \to 0}$ in measure on $\mathbb{R}$ for every ${\alpha > 0.}$

A sequence of functions on $(2, +\infty)$#

Example 15. On ${(2, +\infty)}$ with the Lebesgue measure ${\lambda,}$ consider the functions ${f_n \ge 0}$ given by

$$f_n(x) = \frac{n^2 + 1}{n^2 (x^2 - 2) + (n^2 + 1) x}, \qquad x \in (2, +\infty).$$

(1) What is the pointwise limit of the functions $f_n$ of Example 15? For every ${x \in (2, +\infty),}$

$$\begin{aligned} f_n(x) &= \frac{n^2 (1 + 1/n^2)}{n^2 (x^2 - 2 + x + x/n^2)} \\ &= \frac{1 + 1/n^2}{x^2 - 2 + x + x/n^2} \\ &\xrightarrow[n \to \infty]{} \frac{1}{x^2 + x - 2} = \frac{1}{(x - 1)(x + 2)} =: f(x), \end{aligned}$$

since ${1/n^2 \to 0}$ and ${x/n^2 \to 0.}$

(2) Does $(f_n)$ converge in ${L^1((2, +\infty))?}$ We have

$$\begin{aligned} |f_n| = f_n &= \frac{1 + 1/n^2}{x^2 - 2 + x + x/n^2} \\ &\le \frac{2}{x^2 - 2 + x + 0} \\ &= \frac{2}{x^2 + x - 2} =: g(x), \end{aligned}$$

since ${1/n^2 \le 1}$ and ${x/n^2 \ge 0.}$ Moreover, ${g \in L^1((2, +\infty)),}$ since ${g(x) \sim 2/x^2}$ as ${x \to +\infty.}$ Hence, by the dominated convergence theorem, Theorem 8 with ${p = 1,}$ ${f_n \to f}$ in ${L^1((2, +\infty)).}$

(3) Compute ${\lim_n \int_2^\infty f_n \, d\lambda.}$

Remark 16. Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${f_n, f \colon X \to \mathbb{R}}$ be measurable, for every ${n.}$ If ${f_n \to f}$ in ${L^1,}$ then ${\lim_n \int_X |f_n| \, d\mu = \int_X |f| \, d\mu.}$

By Remark 16, applied to the functions ${f_n \ge 0}$ and ${f \ge 0}$ of (2),

$$\lim_{n \to \infty} \int_2^\infty f_n \, d\lambda = \int_2^\infty \frac{1}{(x - 1)(x + 2)} \, d\lambda.$$

Notice that ${f \ge 0}$ and that ${|f| = f}$ is integrable over $(2, +\infty)$ in the sense of the generalized Riemann integral. Hence, by (2) of Theorem 17 of the fifth post,

$$\int_2^\infty \frac{1}{(x - 1)(x + 2)} \, d\lambda = \lim_{R \to \infty} \int_2^R \frac{1}{(x - 1)(x + 2)} \, dx.$$

Now

$$\begin{aligned} \int_2^R \frac{1}{(x - 1)(x + 2)} \, dx &= \int_2^R \Big( \frac{1/3}{x - 1} + \frac{-1/3}{x + 2} \Big) \, dx \\ &= \frac{1}{3} \Big( \log|x - 1| \, \Big|_2^R - \log|x + 2| \, \Big|_2^R \Big) \\ &= \frac{1}{3} \biggl( \log\Big( \frac{R - 1}{R + 2} \Big) - \log 1 + \log 4 \biggr) \\ &= \frac{1}{3} \log 4 + \frac{1}{3} \log\Big( \frac{R - 1}{R + 2} \Big), \end{aligned}$$

where ${\log 1 = 0.}$ Therefore the limit in (3) is

$$\begin{aligned} &\int_2^\infty \frac{1}{(x - 1)(x + 2)} \, dx \\ &\quad = \lim_{R \to \infty} \biggl( \frac{1}{3} \log 4 + \frac{1}{3} \log\Big( \frac{R - 1}{R + 2} \Big) \biggr) \\ &\quad = \frac{1}{3} \log 4. \end{aligned}$$