There was an exercise, in my professor's book, asking to prove the continuity of an integral depending upon a parameter. Namely, the hypothesis were: Let $D$ be a measurable subset of $\mathbb{R}^n$, and $t_0\in \mathbb{R}$, and let $f:D \times \mathbb{R}\rightarrow \bar{\mathbb{R}}$ such that
- $\exists \lim_{t\to t_0} f(\vec x, t) = g(\vec x)$
- The function $\vec x \mapsto f(\vec x, t)$ is measurable in D $\forall t\in\mathbb{R}$ and exists a summable function $h:D \rightarrow \mathbb{R}$ such that $|f(\vec x, t)|\leq h(\vec x)$ for almost any $\vec x$ in $D$ and $\forall t \in \mathbb{R}$ Then $$\exists \lim_{t \to t_0}\int_D f(\vec x, t) d \vec x = \int_{D} g(\vec x)d\vec x $$
Now, my question is: can't the second hypothesis be eliminated? My reasoning was that if $$\lim_{t \to t_0} f(\vec x, t)=g(\vec x)$$ then $\forall \epsilon >0 \exists \delta > 0:|t-t_0|<\delta \implies |f(\vec x, t)-g(\vec x)|<\frac{\epsilon}{m_n(D)}$, where $m_n(D)$ is the Lebesgue measure of D. But then $$g(\vec x) - \frac{\epsilon}{m_n(D)}<f(\vec x,t)<g(\vec x)+ \frac{\epsilon}{m_n(D)}$$ for $t$ sufficiently close to $t_0$, so $$\int_D g(\vec x)d \vec x - \epsilon<\int_D f(\vec x, t)d \vec x<\int_D g(\vec x)d\vec x + \epsilon$$ and we're finished. Is there something wrong with my reasoning? Thank you guys!