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:
- For every $t\in[0,1]$, the function $x \rightarrow f(x,t)$ is integrable on $[0,1]$ (meaning $\int_{0}^{1}f(x,t)dx < \infty$)
- 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.
- 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!