Lebesgue integration: integral of continuous function tends to infinity

Question

I'm studying measure theory and Lebesgue integration and I've run into this problem:

let $ f:R \rightarrow R$ be a continuous function such that:

• $\int f^+d\lambda_1 = \int f^-d\lambda_1 =+\infty $

show that $c \in R$, then there is $A\in \mathscr{B}(R)$ such that $\int_A fd\lambda_1 = c$.

I know that because of the continuity of $f$, there must be an $A$ such that $0 <\int_A fd\lambda_1 <\infty $. I was thinking of a proof along the lines of the proof of the Rienman series theorem, but I am not sure how to proceed. Some help would be highly appreciated.

Answer 1

I'll give it a shot.

Consider $A = \{x \in \mathbb{R}, f(x) \geq 0\}$ which is well defined since $f$ is continuous.

Consider now $g(x) = \int_{A \cap (0,x)} f(y) dy + \int_{A \cap (-x,0)} f(y) dy = \int_{-x}^x f(y) \mathbf{1}_A(y) dy$.

Clearly $g(0) = 0$. Moreover, $\lim \limits_{x \to \infty} g(x) = \infty$ by your condition on $f_+$. It only remains to show that $g$ is continuous to use the intermediate value theorem.

Usually, to get $g$ continuous, (see for instance this http://mathonline.wikidot.com/continuity-of-functions-defined-by-lebesgue-integrals), you would need $f_+$ to be bounded by an integrable function, but in fact since $f$ is continuous, on any compact set you have that $f_+$ is bounded (Weierstrass theorem). This is enough for your purpose: for any $c > 0$, it exists $x_c > 0$ such that $g(x_c) > c$ (otherwise the limit of $g$ can not be $+ \infty$). Then on the compact $[-x_c - 1, x_c + 1]$, your function $g$ is clearly continuous (since then $f_+$ is bounded) by applying Theorem 1 in the link above.

So, $g(0) = 0, g(x_c) > c$ and g is continuous, therefore it exists $\alpha$ such that $g(\alpha) = c$ (intermediate value theorem https://en.wikipedia.org/wiki/Intermediate_value_theorem). You can do the exact same thing for $c < 0$ by changing the definition of $A$.

Answer 2

Consider $A:=\{x:f^+(x)>0\}$ and $B:=\{x:f^-(x)>0\}$. If $c>0$ then define a function $g:[0,\infty)\rightarrow [0,\infty)$ such that $g(r)=\int_{[-r,r]}f^+(x)dx=\int_{A\cap[-r,r]}f(x)dx.$ Notice that $g(0)=0$ and $g(r)\rightarrow\infty$ as $r\rightarrow \infty.$ And $|g(r)-g(s)|=\int_{[-r,-s)}f^+(x)dx+\int_{(s,r]}f^+(x)dx\rightarrow 0$ as $s\uparrow r.$ So, $g$ is a continuous function. By mean value theorem $\exists b>0$ such that $g(b)=c.$ Take $A_0=[-b,b]\cap A.$ Then $\int_{A_0}f(x)dx=g(b)=c.$ For $c<0$ start with $B$, and for $c=0$ take $A_0=\phi.$