Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I tried to solve the following question while studying for my exams, and since I have no solution for it, I would love it if you could tell me if my proof makes sense. The context is Lebesgue integrals.


Let $f:[0,1]\times[0,1] \rightarrow \mathbb{R}$, where:

  1. For every $t\in[0,1]$, the function $x \rightarrow f(x,t)$ is integrable on $[0,1]$ (meaning $\int_{0}^{1}f(x,t)dx < \infty$)
  2. For every $(x,t) \in [0,1]\times[0,1]$, the partial derivative: $$ g(x,t) := \frac{\mathrm{d}f}{\mathrm{d} t}(x,t) $$ exists.
  3. And finally: $$ M:= sup\left \{|g(x,t)| : 0\leq x,t\leq1 \right \} < \infty $$

Show that $$ F(t) := \int_{0}^{1} f(x,t)dx $$ is derivable in [0,1], and that $$ F'(t) = \int_{0}^{1}g(x,t)dx $$

My Solution

Let $$g_n(x,t) := \frac{f(x, t + \frac{1}{n}) - f(x,t)}{\frac{1}{n }}.$$ So according to (2), $\underset{n \rightarrow \infty}{lim} g_n = g(x, t)$, and from (3), there exists some $M'$ such that for large enough $n$, all $g_n$ are bounded by the same $M'$.


$$ F'_n(t) := \frac{\int_{0}^{1} f(x,t + \frac{1}{n})dx - \int_{0}^{1} f(x,t)}{\frac{1}{n}}dx $$

So $F'_n(t) = \int_{0}^{1} g_n(t)dx$, and according to the bounded convergence theorem, $\underset{n \rightarrow \infty}{lim} F'_n(t) = \int_{0}^{1}g(x,t)dx$, and since from (3), $f$ is Lipschitz in $[0,1]\times[0,1]$, and therefor g is integrable for every t, I think we are done.


share|cite|improve this question
+1 for showing your work. I think there is a typo in the first line, it should be $\lim_n g_n(x,t)=g(x,t)$. You can express the bound $M'$ with $M$ thanks to the mean value theorem. In fact $g$ is integrable since it's bounded over a finite measure set. Last thing: integrable means that $\int_{[0,1]}|f(x,t)|<\infty$ (you need the absolute value). – Davide Giraudo Feb 25 '12 at 11:36
one more thing is that you have to work with a general $t_n\to 0$, not just $1/n$. – Ashok Feb 25 '12 at 11:59
@DavideGiraudo Thanks! In fact we defined integrable functions without the absolute value, and then proved that f is integrable iff |f| is, but since it's the second time I get this comment, I may start using |f| online instead ;). – Hila Feb 25 '12 at 12:02
@Ashok Now that you say it, it makes sense, but all the proofs in my notebook work with $\frac{1}{n}$, so now I'm confused. Did my professor leave something out? – Hila Feb 25 '12 at 12:06
@Hila You may use 1/n or any other sequence the important is the $g_n$ are converging to the properly limit. The bound of the $g_n$ cannot come from the mean value since the derivatives derivatives are only almost everywhere (a.e.). I think in exam you should explain better this bound since it is crucial. – checkmath Feb 25 '12 at 12:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.