Let $f$ be a continuous function on $(0,1]$ and is defined as $f: [0,1] \to \mathbb R$. Show that if $f$ is lebesgue integrable on $[0,1]$, the improper Riemann integral $\lim_{\epsilon \to 0} \int_{\epsilon}^1 f(x)\,dx$ exists and it equals the Lebesgue integral on $[0,1]$.
My guess:
I have read somewhere that Lebesgue and Riemann integral are equal when the function is continuous and the interval is compact. Is it useful in my case?
Best Answer
Your assumption "Show that if $f$ is Lebesgue-integrable" amounts to assuming that $$ \int_{(0,1]} |f| < \infty. $$ Suppose first that $f$ is positive. Then the function $$ g(x)= \int_{[x,1]} f $$ (with the integral being the Lebesgue integral) is a continuous function of $x$ and is bounded above by assumption. Since $g(x)$ increases as $x$ decreases towards $0$, the limit $$ \lim_{x \searrow 0^+} g(x) = \int_{(0,1]}f $$ exists and is equal to the Lebesgue-integral of $f$, because $$ \int_{(0,x)} f \longrightarrow 0 $$ is the difference between $g(x)$ and the integral of $f$ over $(0,1]$ and it goes to zero. Since the Riemann integral over the interval $[x,1]$ and the Lebesgue integral over the same interval coincide, the improper integral exists in this case.
To complete the proof, write $$ f_+ = \max \{f,0\}, \qquad f_- = \max\{-f,0\}. $$ Notice that $f = f_+ - f_-$, and since both functions are positive and continuous you can apply the result above. By the linearity of both integrals (Riemann and Lebesgue) over sums of continuous functions, you are done.
Hope that helps,