Let $f: [c, \infty) \to [0, \infty)$ be bounded function. Suppose $I:=\displaystyle\int_c^{\infty} f(x) dx$ exists.

Then, prove that

(i) $f$ is integrable on $[c, \infty)$ i.e. $f$ is Lebesgue-measurable and $\displaystyle\int_{[c,\infty)} f(x) dx<\infty$.

(ii) $I=\displaystyle\int_{[c,\infty)} f(x) dx.$

For (ii), I tried and it seems to work.

Since $f$ is bounded, there exists $M>0$ s.t. $|f|\leqq M.$

Let $g_n(x):=f(x)\chi_{[c,n]} (x).$

Since $\displaystyle\lim_{n\to \infty} g_n(x)=f(x)\chi_{[c,\infty)} (x)$ and $|g_n(x)|\leqq M,$ from the dominated convergence theorem, I get $\displaystyle\lim_{n\to \infty} \displaystyle\int g_n(x)dx=\int f(x)\chi_{[c,\infty)} (x)dx=\int_{[c, \infty)} f(x) dx.$

But I couldn't proceed more. And for (i), I have no idea what to do.

I'd like you to give me some ideas.

  • $\begingroup$ Riemann integrable functions on a bounded interval $[a,b]$ are Lebesgue integrable. Positive improper integrable functions over infinite intervals $[c,\infty)$ are Lebesgue integrable on $[c,\infty)$. Truncating as you did, gives you a sequence of integrable function that converge monotonically to your target function. the conclusion follows from monotone convergence. $\endgroup$
    – Mittens
    Jun 19, 2021 at 14:42

2 Answers 2


Answer for (i): Let $d >c$. On $[c,d]$ $f$ is Riemann integrable. This implies that it is bounded and continuous almost everywhere. [See Lebesgue_Vitali Theorem in https://en.wikipedia.org/wiki/Riemann_integral ] This is true for each $d>c$ so $f$ is continuous almost everywhere on $[c,\infty)$ and this implies that it is Lebesgue meaurable. Since $f\geq 0$ and $\sup_{d>c} \int_c^{d} f(x)dx <\infty$ it follows that $\int_c^{\infty} f(x)dx <\infty$.

You can complete (ii) by just recalling that $\int_c^{\infty} f(x)dx$ is defined as the limit of $\int_c^{n} f(x)dx$. (The Lebesgue integral of $g_n$ is nothing but the Riemann integral of $f$ from $c$ to $n$).


(i) If $f$ is Riemann integrable over the bounded interval $[a,b]$ then $f$ is Lebesgue integrable:

A simple way to see this is to consider upper and lower Darboux summations. For any $n\in\mathbb{N}$ there is a partition $P_n=\{a=x_{n,0}<\ldots x_{n,m_n+1}=b\}$ such that $$ U(f,P_n)-L(f,P_n)<\frac1n$$ Notice that $$\begin{align} S_n&=\sum^{m_n}_{k=1}M_{n,k}\mathbb{1}_{(x_{n,k-1},x_{n,k}]}\\ s_n&=\sum^{m_n}_{k=1}m_{n,k}\mathbb{1}_{(x_{n,k-1},x_{n,k}]} \end{align}$$ where $M_{n,k}=\sup_{x\in[x_{n,k-1},x_{n,k}]}f(x)$ and $m_{n,k}=\inf_{x\in[x_{n,k-1},x_{n,k}]}f(x)$, are step functions such that $s_n\leq f \leq S_n$ The Riemann integrability condition in terms of upper and lower sums shows that $s_n$ and $S_n$ converge almost surely to $f$ (some details need to be addressed) from where you see that $f$ is Lebesgue measurable on $[a,b]$ Furthermore, the Riemann integral and the Lebesgue integral coincide: $\int^b_as_n\leq\int^a_bf\leq \int^b_aS_n$; Riemann integrals of step functions are exactly the same to their Lebesgue integrals.

(ii) Once measurability of $f$ has been established over any bounded interval, then the functions you define $$g_n=f\mathbb{1}_{[c,n]}$$ are Lebesgue measurable and $g_n$ converge monotonically to $f$ everywhere; hence $f$ is Lebesgue measurable on $[c,\infty)$. The conclusion then follows by monotone convergence.

If Lebeshe measurability is of some cancer, recall that for any Lebesgue measurable function $f$ there is a Borel measurable function $f_0$ such that $f=f_0$ almost everywhere. Substituting $f$ by $f_0$ does not change the value of the integrals.


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