# Improper Riemann integral is equal to Lebesgue integral.

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.

• 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. Jun 19, 2021 at 14:42

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.