# improper Riemann integral and Lebesgue integral

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?

• It doesn't make sense to integrate $f$ over $[0,1]$ if $f$ is defined over $(0,1]$. If you meant that $f : [0,1] \to \mathbb R$ but that $f$ is continuous on $(0,1]$ then you should have said so, it's a different statement. – Patrick Da Silva Oct 21 '12 at 23:16
• So the point here is that $f$ could be discontinuous at $0$ so that the "what you read somewhere" does not apply to $f$ itself. Of course the two integrals on intervals $[a,1]$ where $a>0$ agree. But then you need to do some more work. – GEdgar Oct 21 '12 at 23:20

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.