4
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Oct 21, 2012 at 23:16
  • $\begingroup$ 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. $\endgroup$
    – GEdgar
    Commented Oct 21, 2012 at 23:20

1 Answer 1

3
$\begingroup$

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,

$\endgroup$
2
  • 1
    $\begingroup$ Sorry to bother you after 9 years, but how do we know that $g(x)$ is continuous? In particular, how do we know that $g(x)$ is continuous at $x = 0$? $\endgroup$
    – Leonidas
    Commented Dec 1, 2021 at 19:13
  • $\begingroup$ @Leonidas : We don't know anything about $g$ at $x=0$ a priori, but we don't need to. But by the fundamental theorem of calculus, $g$ is differentiable and its derivative is $f$, and therefore it is continuous. That's what I presume I implicitly used 9 years ago, but I was a young man back then! The integrability of $f$ implies that the integral of $f$ over $]0,x[$ goes to zero with $x$; you can see this clearly when $f$ is a simple function (since then $\int_{(0,x)} f \le ax$ for $x$ small enough) and you can apply the monotone convergence theorem to prove it in the positive measurable case. $\endgroup$ Commented Dec 2, 2021 at 21:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .