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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.

Question

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 $M'$.

Let

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

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.

Thanks!

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    +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).2012-02-25
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    one more thing is that you have to work with a general $t_n\to 0$, not just $1/n$.2012-02-25
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    @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 ;).2012-02-25
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    @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?2012-02-25
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    @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.2012-02-25

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