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Good day!

Need to prove the existence of $t\in [0,1]$ such that $$\int\limits_{0}^{t} f(x)\,dx = \frac{1}{2} \int\limits_{0}^{1} f(x)\,dx,$$ where $f$ is integrable.

My solution: $$F(t)=\int\limits_{0}^{t} f(x)\,dx-\frac{1}{2}\left(\int\limits_{0}^{t} f(x)\,dx +\int\limits_{t}^{1} f(x)\,dx\right)=\frac{1}{2} \int\limits_{0}^{t} f(x)\,dx-\frac{1}{2} \int\limits_{t}^{1} f(x)\,dx.$$

$$F(0)=- \int\limits_{0}^{1} f(x)\,dx,$$ $$F(1)= \int\limits_{0}^{1} f(x)\,dx.$$

If $F(t)$ is continuous than o.k.(Intermediate Value Theorem). It maybe a silly question, but.. if $f$ is integrable function, is a primitive for this function continuous?

I can't imagine a counterexample. And don't know what to do if $\int\limits_{0}^{1} f(x)\,dx=0$.Thanks.

Michael Hardy
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Vasili_Petrov
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3 Answers3

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$F(t)$ is always continuous. This is because $$ F(t+h) - F(t) = \int_0^1 \mathbb I_{[t,t+h]}(x) f(x) \, dx $$ where $\mathbb I_{[t,t+h]}(x)$ is $1$ if $t \le x \le t+h$ and $0$ otherwise. Since $\mathbb I_{[t,t+h]}(x) f(x) \to 0$ when $h \searrow 0$ and $f$ is integrable, by the dominated convergence theorem, $F(t+h) - F(t) \to 0$ when $h \searrow 0$. Similarly if $h \nearrow 0$.

Since $F(0) = 0$ and $F(1) = \int_0^1 f(x) \, dx$, by the IVT there exists $t \in [0,1]$ such that $F(t) = \frac 12 F(1)$ because $\frac 12 F(1)$ is between $F(1)$ and $F(0)$. If the integral is zero you can take $t=0$ or $t=1$ by the way.

Hope that helps,

Patrick Da Silva
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We assume that by integral you mean the Riemann Integral. Then $F(t)=\int_0^tf(x)\,dx$ is a continuous function of $t$ for any integrable $f$.

Without loss of generality we may assume that $\int_0^1 f(x)\,dx =I\ge 0$.

If $I=0$, let $t=0$.

Otherwise, by the Intermediate Value Theorem, for any $b$ such that $0\le b\le I$, there is a $t$ such that $F(t)=b$. Let $b=\frac{I}{2}$.

André Nicolas
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  • I was into measure theory when I answered the question (in the proof of existence/unicity of the Haar measure to be precise, so it was pretty intense) ; I didn't even think it could've been the Riemann integral... haha :) +1 – Patrick Da Silva Jul 03 '13 at 06:05
  • @PatrickDaSilva: Well, I was assuming, or more precisely guessing, there was not enough information to tell. Perhaps it is the Lebesgue integral. – André Nicolas Jul 03 '13 at 06:11
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If $f$ is integrable on a $[a,b]$ then its primitive is uniformly continuous! (Lipschitz continuous to be more precise)

If $f$ is integrable (In Riemann sense) on $[a,b]$, then is also bounded on [a,b], then: $$ |F(y)-F(x)|=\Big |\int_a^y f(t) \, dt -\int_a^x f(t) \, dt \Big| \\=\Big |\int_x^y f(t) \, dt \Big| \le M |y-x| $$

I always liked this proof, so I may as well share it.

matias
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